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"source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "source_path": "/home/ubuntu/upload/25数论训练营综合测试解析版(1).pdf", "sha256": "0fa2def0829c02d28a6f5d88597b6bfc4a369cfd62cb9b0ea3fa28471d11ea97", "section_type": "test", "lesson_or_test": "数论训练营 综合测试", "total_pages": 2, "question_count": 18, "page_notes": {}, "source_notes": "", "topics_in_source": [], "question_ids": [ "S14-Q001", "S14-Q002", "S14-Q003", "S14-Q004", "S14-Q005", "S14-Q006", "S14-Q007", "S14-Q008", "S14-Q009", "S14-Q010", "S14-Q011", "S14-Q012", "S14-Q013", "S14-Q014", "S14-Q015", "S14-Q016", "S14-Q017", "S14-Q018" ] } ], "questions": [ { "id": "S01-Q001", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "例题1", "problem": "已知十位数\\overline{1a2a3a4a5a}能被11整除，则a=____。", "source_answer": "3", "source_solution": "11|(5a−15)⇒11|5(a−3)⇒11|(a−3)⇒a=3。", "topics": [ "整除特征：11的整除法则", "数字谜" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "3", "solution_status": "consistent", "note": "奇偶位数字和之差为15−5a，a为数字时仅a=3使其为11的倍数。", "verified_solution": "十位数1a2a3a4a5a的奇数位数字和是1+2+3+4+5=15，偶数位数字和是5a。能被11整除，要求两组数字和之差15−5a是11的倍数。a是0到9的数字；逐个看或解这个范围内的倍数方程，只有15−5a=0时成立，所以a=3。" }, "cross_reference_ids": [] }, { "id": "S01-Q002", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "练习1", "problem": "2026年，疯狂动物城警察局为了表彰表现优异的警察，给他们授予特殊的警号。这些警号是形如\\overline{a2026b}的六位数，并且都能被63整除，这样的警号有那些？", "source_answer": "120267，820260", "source_solution": "63=7×9；由9的整除性得a+b=8或17。结合7的整除性，a+b=8时a=1,8，得到120267、820260；a+b=17无解。", "topics": [ "整除特征：9的整除法则", "整除特征：7的整除检验", "数字谜" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "120267、820260", "solution_status": "consistent", "note": "原页明确写作a2026b；题库旧题干转录把数字顺序写成2026ab，已按原页修正。独立检验确认答案120267、820260。", "verified_solution": "原题的数是a2026b，a是首位数字（1到9），b是末位数字（0到9）。63=7×9。能被9整除要求a+b+10是9的倍数，因此a+b=8或17。又a2026b除以7的余数与5a+b+2相同。若a+b=8，则5a+b+2=4a+10，得到a=1或8，分别是120267和820260。若a+b=17，则得到a≡4 (mod7)，但a只能为1到9，且b=17−a也必须是0到9；唯一可能的a=4会使b=13，不是数字。因此只有120267和820260。题库原转录把数字顺序写成2026ab，现按原页改为a2026b。" }, "cross_reference_ids": [] }, { "id": "S01-Q003", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "例题2", "problem": "(1)在算式\\overline{1□9}×\\overline{9□7}×\\overline{7□1}的空格中各填入一个数字，使得最后的乘积能被99整除。那么所填的三个数字之和的最大值与最小值之差（大减小）是多少？(2)已知七位数\\overline{14xy78z}能被792整除，则\\overline{xyz}等于____。", "source_answer": "(1)19；(2)664", "source_solution": "(1)99=9×11。若11|\\overline{9b7}则b=5，3|\\overline{1a9}×\\overline{7c1}；若11|\\overline{7c1}则c=8，9|\\overline{1a9}×\\overline{9b7}。原解写出两组极值：(a+c)max=17、(a+c)min=1；(a+b)max=17、(a+b)min=2，故(17+8)−(1+5)=19。(2)792=8×99；由8的规则得z=4，由99的规则得x=y=6，所以\\overline{xyz}=664。", "topics": [ "整除特征：8的整除法则", "整除特征：9的整除法则", "整除特征：11的整除法则", "数字谜" ], "page_refs": { "problem": [ 1, 2 ], "answer": [ 1, 2 ], "solution": [ 1, 2 ] }, "ambiguities": [ "第1问原题三个空格未标字母，解析中以a、b、c代替；不影响题意。" ], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "19", "solution_status": "consistent", "note": "穷举三数字可得所填数字和范围为6至25，差19；664对应七位数1466784，确被792整除。", "verified_solution": "设三个空格依次为a、b、c。99=9×11。11不能整除1a9，因为它的奇偶位数字和之差在1到10之间；因此11必须整除9b7或7c1。前者要求b=5，后者要求c=8。再用9的倍数条件：若b=5，至少一个数1a9、7c1能被3整除，分别要求a+10或c+8是3的倍数。此时a+b+c最小为6（a=0、c=1），最大为22。若c=8，至少一个数1a9、9b7能被3整除；对应a+10或b+16是3的倍数，此时和最小为10，最大为25。总的最小和为6，最大和为25，差为25−6=19。" }, "cross_reference_ids": [] }, { "id": "S01-Q004", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "练习2", "problem": "一个六位数\\overline{a1991b}能被12整除，求这样的六位数的个数。", "source_answer": "6个", "source_solution": "4|\\overline{a1991b}⇒4|\\overline{1b}⇒b=2,6；分别结合3的整除性，b=2时a=2,5,8，b=6时a=1,4,7，共3+3=6个。", "topics": [ "整除特征：3的整除法则", "整除特征：4的整除法则", "计数" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "6个", "solution_status": "consistent", "note": "12=3×4，列出的6个a,b组合均满足两条件。", "verified_solution": "12=3×4，且3和4互质，所以数要同时被3和4整除。末两位是1b，能被4整除时b只能是2或6。各位数字和为a+b+20。若b=2，要被3整除就需a+22是3的倍数，a=2、5、8，共3种；若b=6，需a+26是3的倍数，a=1、4、7，也有3种。总数为6个。" }, "cross_reference_ids": [] }, { "id": "S01-Q005", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "例题3", "problem": "已知六位数\\overline{18abc2}可以被17和31整除，则a+b+c的最大值为____。", "source_answer": "14", "source_solution": "527=17×31。检验18abc2范围内的527倍数：189992÷527=360余272，故189992−272=189720；189720−527×4=187612，对应abc=761、a+b+c=14；189720−527×14=182342，对应abc=234、a+b+c=9。因此最大值为14。", "topics": [ "合数分解与整除", "数字谜", "最大值问题" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "14", "solution_status": "consistent", "note": "回看第2页原图后确认手写数为187612与182342；它们分别对应数字和14、9，来源答案14正确。此前片段把手写数误读，现已纠正。", "verified_solution": "17和31互质，所以同时被两者整除等价于被527=17×31整除。六位数范围是180002到189992，且末位固定为2。189992除以527余272，因此低于它的最近一个527倍数是189720。后续倍数写作189720−527t；末位为2要求7t≡8 (mod10)，所以t≡4 (mod10)。在保持首两位为18的范围内只有t=4、14，得到187612、182342。它们的a+b+c分别为14和9，最大值14。" }, "cross_reference_ids": [] }, { "id": "S01-Q006", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "练习3", "problem": "六位自然数\\overline{1082□□}能被23整除，末两位数有几种情况？", "source_answer": "4种：84，59，36，13", "source_solution": "108299÷23=4708…15；末两位从00至99变化时余数相应变化，列出可行末两位84、59、36、13，共4种。", "topics": [ "整除与余数", "数字谜", "计数" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "partially_incorrect", "verified_answer": "末两位15、38、61、84，共4种。", "solution_status": "error", "note": "来源所列84、59、36、13中只有84能被23整除；正确末两位是15、38、61、84，共4种。因此来源给出的数量正确，具体清单错误；已保留原文并记录正确列表。", "verified_solution": "设末两位组成的数为x，0≤x≤99。108200除以23余8，所以108200+x能被23整除，等价于x除以23余15。范围内这样的数是15、38、61、84，共4种。逐个对应的六位数为108215、108238、108261、108284。原卷列出的84、59、36、13中，只有84正确；答案“共4种”正确，但数字清单错误。" }, "cross_reference_ids": [] }, { "id": "S01-Q007", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "例题4", "problem": "如果653整除\\overline{ab2347}，则a+b=____。", "source_answer": "11", "source_solution": "方法一作竖式/倍数检验；方法二：653|\\overline{ab2347}+653，化为653|\\overline{ab3000}，再化为653|\\overline{ab3}，故\\overline{ab3}=653，a=6,b=5。", "topics": [ "整除与倍数", "数字谜" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "11", "solution_status": "consistent", "note": "原页视觉确认题干为\\overline{ab2347}。由N+653=1000×\\overline{ab3}且(653,1000)=1，可得\\overline{ab3}=653；故a=6、b=5，核验652347=653×999。", "verified_solution": "把六位数写成N=1000×(100a+10b+3)+347? 更直接按原页构造：N=\\overline{ab2347}，N+653=\\overline{ab3000}=1000×\\overline{ab3}。若653|N，则653也整除N+653；又653与1000互质，所以653|\\overline{ab3}。\\overline{ab3}是三位数，只可能等于653，因此a=6、b=5，a+b=11。检查652347=653×999。" }, "cross_reference_ids": [] }, { "id": "S01-Q008", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "练习4", "problem": "一个自然数的末两位数字为17，它的数字和为17，且能被17整除，满足条件的最小五位自然数是____。", "source_answer": "15317", "source_solution": "末两位为17且数字和为17，前三位数字和为9；又因17|\\overline{abc17}，化简为17|\\overline{abc}，并结合9的条件，最小的前三位数为153，所以最小五位数为15317。", "topics": [ "整除特征：17的整除检验", "数字和", "最小值问题" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "15317", "solution_status": "consistent", "note": "15317÷17=901，且1+5+3+1+7=17；按前三位递增搜索无更小可行数。", "verified_solution": "五位数可写成100×\\overline{abc}+17。若它能被17整除，由于100与17互质，\\overline{abc}也必须被17整除。数字和条件给出a+b+c=9。三位数的17倍从102、119、136、153……开始；前三个数字和分别为3、11、10，153是第一个数字和为9的数。因此最小五位数是15317；它的数字和是17且15317=17×901。" }, "cross_reference_ids": [] }, { "id": "S01-Q009", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "例题5", "problem": "正整数n使得4n−1整除2024n，求n的值。", "source_answer": "3，6", "source_solution": "(4n−1)|[506(4n−1)+506]，故(4n−1)|506=2×11×23；列正因数并满足4n−1的形式，得到n=3,6。", "topics": [ "带参数整除", "因数分解" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "n=3、6", "solution_status": "consistent", "note": "506的正因数中只有11、23可写成4n−1（分别给n=3,6）。", "verified_solution": "令d=4n−1。因为n为正整数，d是正数，且d|2024n。又2024n=506(4n−1)+506，所以d|506。506=2×11×23，其正因数中余数为3 (mod4)的只有11和23，因为d=4n−1。于是4n−1=11或23，得到n=3或6；两者代回都成立。" }, "cross_reference_ids": [] }, { "id": "S01-Q010", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "练习5", "problem": "使得分式(6x+3)/(2x−1)的值为正整数的自然数x的值有____个。", "source_answer": "2个", "source_solution": "(6x+3)/(2x−1)=3+6/(2x−1)，故2x−1为6的正因数，取1,3，得x=1,2，共2个。", "topics": [ "整除与分式", "因数与计数" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "2个（x=1、2）", "solution_status": "consistent", "note": "x=1、2，共2个；分式值分别是9、5。此前复核备注把x=2的值误写为7，现已更正；原卷解法给出的两个x值正确。", "verified_solution": "当x为正整数时，(6x+3)/(2x−1)=3+6/(2x−1)。要使结果为正整数，奇数2x−1必须是6的正因数；它只能是1或3，所以x=1或2，共2个。代入原式分别得到9和5。即使把0算作自然数，x=0时原式为−3，也不符合正整数要求。" }, "cross_reference_ids": [] }, { "id": "S01-Q011", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "例题6", "problem": "求最大的正整数n，使得n²+100能被n+10整除。", "source_answer": "190", "source_solution": "(n+10)|[(n−10)(n+10)+200]，故(n+10)|200；最大正因数n+10=200，得n=190。", "topics": [ "带参数整除", "因数分解", "最大值问题" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "190", "solution_status": "consistent", "note": "n+10为200的正因数，最大取200；190²+100=36200可被200整除。", "verified_solution": "n²+100=(n+10)(n−10)+200。因此n+10必须是200的正因数。n为正整数，故n+10>10；要让n最大，就取最大的可能因数200，得到n=190。检查190²+100=36200，确实能被200整除。" }, "cross_reference_ids": [] }, { "id": "S01-Q012", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "拓展1", "problem": "已知：(n+42)|42n+1，求所有自然数n的和。", "source_answer": "1722", "source_solution": "(n+42)|[42(n+42)−42²+1]，故(n+42)|(42²−1)=41×43；n+42=43或1763，得n=1或1721，和为1722。", "topics": [ "带参数整除", "因数分解", "求和" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "1722", "solution_status": "consistent", "note": "自然数按正整数理解；两值均直接满足整除条件。", "verified_solution": "令d=n+42。因为42n+1=42(n+42)−(42²−1)，若n+42能整除42n+1，就必须整除42²−1=1763=41×43。n为正整数，所以d≥43；1763的因数中不小于43的只有43和1763。于是n=1或1721，所有n的和为1722。" }, "cross_reference_ids": [] }, { "id": "S01-Q013", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "拓展2", "problem": "已知正整数a满足192|a³+191，且a<2026，求满足条件的所有可能的正整数a的和。", "source_answer": "10571", "source_solution": "192|(a³+192−1)⇒192|(a−1)(a²+a+1)。因a²+a+1=a(a+1)+1与a互质，先得64|(a−1)，设a=64k+1；再化简得3|k，令k=3m，m=0,1,…,10，故a=192m+1，和=192(1+…+10)+11=10571。", "topics": [ "同余与整除", "因式分解", "周期性与计数", "求和" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "10571", "solution_status": "consistent", "note": "独立穷举1≤a<2026得到1,193,385,…,1921共11项，和10571。", "verified_solution": "由192|(a³+191)可知a³≡1 (mod192)，所以分别有64|(a³−1)和3|(a³−1)。因a³−1=(a−1)(a²+a+1)，而a²+a+1总是奇数，64必须整除a−1，因此a=64k+1。再看模3，a³≡a，所以a≡1 (mod3)；代入a=64k+1得k≡0 (mod3)。于是a=192m+1。条件1≤a<2026给出m=0,1,…,10，11个数的和为11+192(0+1+…+10)=10571。" }, "cross_reference_ids": [] }, { "id": "S01-Q014", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "作业1", "problem": "一个四位数\\overline{a58b}，能同时被5和9整除，那么这个数最大是____。", "source_answer": "9585", "source_solution": "被5整除时b=0或5；被9整除时，b=0得a=5，b=5得a=9，最大数为9585。", "topics": [ "整除特征：5的整除法则", "整除特征：9的整除法则", "最大值问题" ], "page_refs": { "problem": [ 6 ], "answer": [ 6 ], "solution": [ 6 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "9585", "solution_status": "consistent", "note": "5580与9585均满足条件，比较得最大值9585。", "verified_solution": "四位数a58b要被5整除，末位b只能是0或5。各位数字和a+5+8+b必须是9的倍数。b=0时a=5，得到5580；b=5时a=9，得到9585。比较两数，最大是9585。" }, "cross_reference_ids": [] }, { "id": "S01-Q015", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "作业2", "problem": "四位数32□□能被23整除，则这个四位数最大是____，最小是____。", "source_answer": "最大3289，最小3220", "source_solution": "3299÷23=143余10；从3299向下找23的倍数，依次为3289、3266、3243、3220。结合四位数32□□的范围，最大3289，最小3220。", "topics": [ "整除与余数", "最大值与最小值" ], "page_refs": { "problem": [ 6 ], "answer": [ 6 ], "solution": [ 6 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "最大3289，最小3220", "solution_status": "consistent", "note": "原页红笔答案与候选列表均显示3289、3266、3243、3220；旧题库转录把3289误写为3299，现已修正source_answer和source_solution。3299÷23余10，故最大倍数是3289。", "verified_solution": "四位数32□□在3200到3299之间。3299除以23余10，所以不超过3299的最大23倍数是3299−10=3289。依次减23得到3266、3243、3220；再减23会低于3200。因此最大3289、最小3220。原页手写答案是3289和3220；题库旧转录把最大值误记为3299，已按图像修正。" }, "cross_reference_ids": [ "XR-01" ] }, { "id": "S01-Q016", "source_id": "S01", "source_filename": "26数论模块训练营第一讲解析.pdf", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "source_question_label": "作业3", "problem": "若整数A使得(A−42)能整除(42×A−1)，那么这样的A有哪些？", "source_answer": "43，83，85，1805", "source_solution": "(A−42)|[42(A−42)+42²−1]，故(A−42)|41×43；取正因数1,41,43,41×43，得A=43,83,85,1805。", "topics": [ "带参数整除", "因数分解" ], "page_refs": { "problem": [ 6 ], "answer": [ 6 ], "solution": [ 6 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "incomplete", "verified_answer": "按题面“整数 A”，全部解为 A=−1721、−1、1、41、43、83、85、1805。若原意限定 A>0，则取其中六个正整数。", "solution_status": "error", "note": "原页题干明确写“整数A”，现按全部整数作答。令d=A−42，则d|1763=41×43；正负因数都要纳入，完整解为−1721、−1、1、41、43、83、85、1805。来源仅列四个正因数对应的解，漏解。若原意只限正整数，需在题面补明。", "verified_solution": "令 d=A−42（d≠0）。原条件等价于 d | 42(d+42)−1=42d+1763，因此 d |1763=41×43。整数因数为 ±1、±41、±43、±1763；逐一代入 A=42+d，得到 −1721、−1、1、41、43、83、85、1805。原题写“整数”，所以负解也须列出；若教材原意为正整数，应补条件。" }, "cross_reference_ids": [] }, { "id": "S02-Q001", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "例题1（1）", "problem": "把三个最小的连续的奇质数，分别写在三张纸片上，每张纸片上写且仅写一个奇质数，如果随意从其中取出至少一张组成一个数，那么质数有（ ）个。", "source_answer": "6", "source_solution": "三个数为3、5、7；组成的质数列为3、5、7、53、37、73，共6个。", "topics": [ "质数判定", "排列组合与构造" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "6个", "solution_status": "consistent", "note": "单个数和两位数逐一检验，37、53、73为质数，其余两位排列15、35、57等合数。", "verified_solution": "三张纸片上的数是3、5、7。取一张时得到3、5、7，都是质数。取两张并按不同顺序组成两位数，六种是35、37、53、57、73、75；其中37、53、73是质数。取三张组成三位数时，数字和是3+5+7=15，能被3整除，所以都不是质数。总共有3+3=6个质数。" }, "cross_reference_ids": [] }, { "id": "S02-Q002", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "例题1（2）", "problem": "若两个质数 p，q 满足 3p²+5q=517，则 p+q=____。", "source_answer": "103或15", "source_solution": "p、q中必有偶数。p=2时，3×2²+5q=517，q=101；q=2时，3p²+5×2=517，p=13。", "topics": [ "质数奇偶性", "方程与整除" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "103或15", "solution_status": "consistent", "note": "两组分别为(p,q)=(2,101),(13,2)，和为103、15；奇质数同时代入左式为偶数，不可能等于517。", "verified_solution": "如果p和q都是奇质数，那么3p²和5q都是奇数，和是偶数，不可能等于517。因此p或q至少一个等于2。若p=2，3×2²+5q=517，得到q=101，p+q=103。若q=2，3p²+10=517，得到p²=169、p=13，p+q=15。两组都满足条件。" }, "cross_reference_ids": [] }, { "id": "S02-Q003", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "练习1", "problem": "若 P 和 P³+5 都是质数，那么 P⁵+5 等于多少？", "source_answer": "37", "source_solution": "P为偶质数，所以P=2；P³+5=13为质数；P⁵+5=2⁵+5=37。", "topics": [ "质数奇偶性", "质数判定" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "37", "solution_status": "consistent", "note": "若P为奇质数，P³+5为偶数且大于2，不是质数，故P=2。", "verified_solution": "若P是奇质数，P³是奇数，P³+5就是大于2的偶数，不是质数。因此P只能是2。此时P³+5=13确实是质数，而P⁵+5=2⁵+5=37。" }, "cross_reference_ids": [] }, { "id": "S02-Q004", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "例题2", "problem": "60以内四个互不相同的质数从小到大排列，构成一个等差数列（相邻两数的差相等），这样的四个数有____种可能情况。", "source_answer": "7", "source_solution": "列出：5,11,17,23；5,17,29,41；5,23,41,59；7,19,31,43；11,17,23,29；17,29,41,53；41,47,53,59，共7种。", "topics": [ "质数数列", "等差数列", "枚举与排除" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "7种", "solution_status": "consistent", "note": "按公差及首项枚举，7组均为互异质数且末项≤60；含2的四项等差数列不可能。", "verified_solution": "四项依次为p、p+d、p+2d、p+3d。它们不能含2：若d为奇数则p+2d为大于2的偶数；若d为偶数则p+d为大于2的偶数。因此四项都是奇质数，d为偶数。逐个检查不超过60的公差，符合条件的只有：d=6时(5,11,17,23)、(11,17,23,29)、(41,47,53,59)；d=12时(5,17,29,41)、(7,19,31,43)、(17,29,41,53)；d=18时(5,23,41,59)。合计7组。" }, "cross_reference_ids": [] }, { "id": "S02-Q005", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "练习2", "problem": "已知a，b，c是3个彼此不同的质数，若a+b×c=37，则a+b−c最大是____。", "source_answer": "32", "source_solution": "c必有偶数，取c=2，则a+2b=37；取b=3,a=31，(a+b−c)max=31+3−2=32。", "topics": [ "质数奇偶性", "整除与最值" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "32", "solution_status": "error", "note": "答案32正确，但来源证明“c必有偶数”不成立：还需检查b=2的情形，且b、c均为奇数时必须另查a=2。完整分类后最大值仍为32，已补独立证明。", "verified_solution": "若b、c都是奇质数，bc是奇数。要使a+bc=37为奇数，a必须是偶质数2；于是bc=35，只有(b,c)=(5,7)或(7,5)，目标值分别为0、4。若b或c等于2，分开检查：c=2时，可行的(a,b)为(31,3)、(23,7)、(11,13)、(3,17)，对应a+b−c为32、28、22、18；b=2时，可行的(a,c)同为(31,3)、(23,7)、(11,13)、(3,17)，目标值为30、18、0、−12。因此最大值是32。来源解法只考虑c=2，漏掉b=2及a=2的情况；补全后可确认最大值不变。" }, "cross_reference_ids": [] }, { "id": "S02-Q006", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "例题3（1）", "problem": "写出10个连续自然数，使它们个个都是合数。", "source_answer": "114～123", "source_solution": "找到121，取114～123均为合数。", "topics": [ "合数判定", "连续整数构造" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [ "第2页例题3（1）红笔答案出现“114～126均是”，与题目要求的10个数量不一致。" ], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "114、115、116、117、118、119、120、121、122、123（共 10 个）。", "solution_status": "error", "note": "114～123恰为10个连续合数，来源答案成立；手写解析另写“114～126”却包含13个数，与题目要求的10个不一致，属于解析端点/计数错误。", "verified_solution": "正确的一组是 114～123，恰好 10 个连续自然数。逐个分解：114=2×57，115=5×23，116=4×29，117=9×13，118=2×59，119=7×17，120=2×60，121=11²，122=2×61，123=3×41，均为合数。原手写“114～126”包含 13 个数，不符合题目数量。" }, "cross_reference_ids": [] }, { "id": "S02-Q007", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "例题3（2）", "problem": "找200个连续的自然数它们个个都是合数。", "source_answer": "201!+2，201!+3，…，201!+201", "source_solution": "取201!+2至201!+201；其中每个数均有对应的2至201作为因子。", "topics": [ "合数判定", "阶乘构造", "连续整数构造" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "201!+2、201!+3、…、201!+201", "solution_status": "consistent", "note": "共有200项；对2≤k≤201，201!+k可被k整除且大于k。", "verified_solution": "取201!+2到201!+201。k从2到201一共有201−2+1=200个整数。201!能被每个这样的k整除，所以201!+k也能被k整除；而201!+k大于k，故它有k和另一个大于1的因数，是合数。" }, "cross_reference_ids": [] }, { "id": "S02-Q008", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "练习3", "problem": "求小于700的最大质数。", "source_answer": "691", "source_solution": "699被3整除，697=17×41；695被5整除，693被3整除，故为691。", "topics": [ "质数判定", "试除法" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "691", "solution_status": "incomplete", "note": "原解列出部分较大合数，但未检查全部692—699，也未证明691是质数；独立解法已补全。", "verified_solution": "699、698、697、696、695、694、693、692都不是质数：偶数直接排除；699、693能被3整除；695能被5整除；697=17×41。691只需检查不超过√691≈26.3的质数2、3、5、7、11、13、17、19、23，没有一个能整除，所以691是小于700的最大质数。" }, "cross_reference_ids": [] }, { "id": "S02-Q009", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "例题4", "problem": "质数P₃是一个三位数，它的数码和为P₂且P₂是一个两位数的质数，而P₂的数码和为P₁且P₁是一个大于2的质数。请问满足上述条件所有可能的P₃之和是什么？", "source_answer": "3260", "source_solution": "P₃=abc⇒P₂≤27⇒P₁≤10，P₁=3,5,7。P₁=3、7无解；P₁=5时P₂=23，P₃为599、977、797、887，共599+977+797+887=3260。", "topics": [ "质数判定", "数码和", "构造与枚举" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "3260", "solution_status": "incomplete", "note": "原解给出四个候选数，但没有把数码和为23的全部三位数列全；已补枚举和筛选。", "verified_solution": "三位数P₃的数码和P₂最多27，所以两位质数P₂只能在10到27内。P₁是大于2的质数，且是P₂的数码和，只能是3、5、7；其中只有P₂=23满足条件。接着枚举数码和为23的三位数：599；689、698；779、788、797；869、878、887、896；959、968、977、986、995。检验这些数后，质数只有599、797、887、977，和为3260。" }, "cross_reference_ids": [] }, { "id": "S02-Q010", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "练习4", "problem": "一个两位数，数字和是质数。而且，这个两位数分别乘以3，5，7之后，得到的数的数字和都仍为质数。满足条件的两位数为____。", "source_answer": "67", "source_solution": "设两位数为P₁=ab。因3|3P₁，3P₁的数码和为3的倍数，必须为3，故3P₁为210、120、102、201、111；P₁候选70、40、34、67、37，验算得67符合。", "topics": [ "数码和", "整除规则", "质数判定" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "与拓展2同文档重复题，题面和答案相同。", "review": { "answer_status": "verified", "verified_answer": "67", "solution_status": "incomplete", "note": "原解筛出候选后没有逐一说明5倍、7倍数码和；已补完整核对，唯一答案67。", "verified_solution": "设两位数为n。3n的数码和是3的倍数，又必须是质数；因3n≤297，数码和只能是3。于是3n只能是102、111、120、201、210，n候选为34、37、40、67、70。逐个检验：34的5倍数码和为8；37和40本身的数码和不是质数；70的5倍数码和为8；67的数码和为13，3倍、5倍、7倍分别为201、335、469，数码和3、11、19都为质数。因此唯一答案是67。" }, "cross_reference_ids": [ "XR-02" ] }, { "id": "S02-Q011", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "例题5", "problem": "小将0、1、2、3、4、5、6、7、8、9这十个数字不重复的填入右侧方格中，横向、竖向相邻的两个方格从左到右、从上到下依次可以组成一个两位数（0不能作为首位），那么，这两位数中，最多有____个质数。", "source_answer": "7", "source_solution": "红笔写“末位只能为1、3、7、9⇒最多7个”，并给出阶梯形填法：顶2；第二行8、4；第三行6、1、9；第四行5、3、7、0。", "topics": [ "质数判定", "数字排列", "图形构造与最值" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [ "题中“横向、竖向相邻的两个方格从左到右、从上到下依次可以组成一个两位数”依赖右侧阶梯方格图；图形在视觉页可辨，文本层不含图。" ], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "7个", "solution_status": "incomplete", "note": "原解未说明“个位只有1、3、7、9”为何给出7条上限；已按原图的入边数补上严格上界与达到上界的填法。题目依赖方格图，需随题保留图形。", "verified_solution": "按原图的阶梯形格子（每行1、2、3、4格）从左到右、从上到下读相邻格。两位质数个位只能是1、3、7、9。图中有3个格子各有2条指向它的相邻边、6个格子各有1条、1个格子没有进入边；四个可能的质数个位最多贡献2+2+2+1=7条质数边，所以不可能超过7。原图填法为：第一行2；第二行8、4；第三行6、1、9；第四行5、3、7、0。它形成的7个质数是41、61、19、13、97、53、37，所以上限能达到，最大就是7。此题必须保留原图或等价格子示意，只有文字题干不完整。" }, "cross_reference_ids": [] }, { "id": "S02-Q012", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "练习5", "problem": "从1−9中选出8个数排成一个圆圈，使得相邻的两数之和都是质数。排好后可以从任意两个数字之间切开，按顺时针方向读这些八位数，其中可以读到的最大的数是多少？", "source_answer": "98567432", "source_solution": "奇偶相间，从高位开始构造；最大为98567432。", "topics": [ "质数奇偶性", "圆排列", "最值构造" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "98567432", "solution_status": "incomplete", "note": "原解给出构造结果，但最大性理由不够完整；已补逐位最大选择证明，并用全部圆圈穷举作独立交叉检查。", "verified_solution": "相邻和为质数时，两个数字必须一奇一偶（相邻数和都大于2，同奇偶会得到偶数）。八个位置绕成圈，因此奇偶交替；1—9中有5个奇数、4个偶数，必须用完四个偶数并少用一个奇数。为让顺时针读出的数最大，先取9；后面依次取能组成质数和的最大未用数字：8（和17）、5（和13）、6（和11）、7（和13）、4（和11）、3（和7），最后剩2，且2+9=11也为质数。因此最大数是98567432。已对所有可能圆圈作穷举复核，最大值一致。" }, "cross_reference_ids": [] }, { "id": "S02-Q013", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "例题6", "problem": "证明：质数有无穷多个。", "source_answer": "质数有无穷多个", "source_solution": "假设质数有限，为p₁n且q≤N−1n。又q≤N−111且为偶数时，n=4+(n−4)，两项均为合数；n>11且为奇数时，n=9+(n−9)，两项均为合数。", "topics": [ "合数判定", "奇偶分类", "构造证明" ], "page_refs": { "problem": [ 5 ], "answer": [ 5 ], "solution": [ 5 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "命题成立", "solution_status": "consistent", "note": "偶数n≥12时n−4≥8为偶合数；奇数n≥13时n−9≥4为偶合数，4和9均合数。", "verified_solution": "若n>11且为偶数，写成4+(n−4)。4是合数，n−4是至少8的偶数，也是假合数。若n为奇数，则n≥13，写成9+(n−9)：9是合数，n−9是至少4的偶数，也是合数。因此每个大于11的整数都能写成两个合数之和。" }, "cross_reference_ids": [] }, { "id": "S02-Q016", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "拓展2", "problem": "一个两位数，数字和是质数。而且，这个两位数分别乘以3，5，7之后，得到的数的数字和都仍为质数。满足条件的两位数为____。", "source_answer": "67", "source_solution": "同练习4，答案写为67。", "topics": [ "数码和", "整除规则", "质数判定" ], "page_refs": { "problem": [ 5 ], "answer": [ 5 ], "solution": [ 5 ] }, "ambiguities": [], "duplicate_notes": "与练习4为同文档重复题，题面几乎完全相同。", "review": { "answer_status": "verified", "verified_answer": "67", "solution_status": "consistent", "note": "同练习4；67满足全部条件。", "verified_solution": "与本讲练习4是同一道题。设两位数为n。3n的数码和既是3的倍数又是质数，只能是3；所以3n只能是102、111、120、201、210，n候选为34、37、40、67、70。逐一核对n及3n、5n、7n的数码和后，只有67满足：13、3、11、19都是质数。" }, "cross_reference_ids": [ "XR-02" ] }, { "id": "S02-Q017", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "作业1", "problem": "已知a,b均为质数，且aᵇ+b=2059，那么ab=____。", "source_answer": "22", "source_solution": "a、b必有2。a=2时，2ᵇ+b=2059，b=11，得ab=22；b>11无解。b=2时a²+2=2059，无解。", "topics": [ "质数奇偶性", "指数方程", "整除与枚举" ], "page_refs": { "problem": [ 7 ], "answer": [ 7 ], "solution": [ 7 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "22", "solution_status": "incomplete", "note": "原解正确指出奇偶性分支，但未检查a=2时较小的质数b=2、3、5、7；已补有限枚举与b>11的单调性界。", "verified_solution": "若a、b都是奇质数，aᵇ是奇数，aᵇ+b为偶数，不可能等于2059。因此a或b为2。若a=2，检查b为质数的可能值：b=2、3、5、7、11时，2ᵇ+b分别为6、11、37、135、2059；更大的质数b会使数值更大，所以只能b=11。若b=2，则a²=2057；因为45²=2025、46²=2116，2057不是平方数。故(a,b)=(2,11)，ab=22。" }, "cross_reference_ids": [] }, { "id": "S02-Q018", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "作业2", "problem": "一个数A为质数，并且A+14，A+18，A+32，A+36也是质数。那A的值是____。", "source_answer": "5", "source_solution": "A=5时19、23、37、41均为质数；A≠5时，A的末位为1、3、7、9、偶数，四个数中必有一个末位为5，不成立。", "topics": [ "质数判定", "模5与末位数字", "排除法" ], "page_refs": { "problem": [ 7 ], "answer": [ 7 ], "solution": [ 7 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "5", "solution_status": "consistent", "note": "答案正确；已将原末位筛选写成模5余数的完整论证。", "verified_solution": "A=5时四个数是19、23、37、41，都是质数。若A不是5，质数A除以5的余数只能是1、2、3、4；四个偏移量14、18、32、36除以5的余数正好是4、3、2、1。无论A余几，总有一个偏移量与它相加后能被5整除，而且这个数大于5，因此不是质数。故只能A=5。" }, "cross_reference_ids": [] }, { "id": "S02-Q019", "source_id": "S02", "source_filename": "26数论模块训练营第二讲解析.pdf", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "source_question_label": "作业3", "problem": "用1、2、3、5、6、7、8、9这8个数字最多可以组成____个质数（每个数字只能使用一次，且必须使用）。", "source_answer": "6", "source_solution": "最多6个；示例为2、3、5、7、61、89，8个数字恰好各用一次。", "topics": [ "质数判定", "数字排列", "最值与构造" ], "page_refs": { "problem": [ 7 ], "answer": [ 7 ], "solution": [ 7 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "6", "solution_status": "incomplete", "note": "来源给出6个质数的构造，但未证明最大性；已补用数字数作上界：7个质数至少需要10个数字，与仅有8个数字矛盾。", "verified_solution": "这8个数字里，能单独成为质数的只有2、3、5、7，一共4个。其他任何质数都至少要用2个数字。假如要拼出7个质数，即使把4个单独的质数全用上，还剩下3个质数必须各用至少2个数字，总共至少要4+2×3=10个数字；如果单独使用的质数更少，需要的数字只会更多。可是我们只有8个数字，所以不可能有7个，最多只能有6个。达到6个的方法是2、3、5、7、61、89：它们全是质数，而且恰好用完1、2、3、5、6、7、8、9。" }, "cross_reference_ids": [] }, { "id": "S03-Q001", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "例题1", "problem": "把下列数分解质因数并写成标准式：（1）2024，2025，999999，142857；（2）101，1001，10001，10101，111，1111，11111，111111。", "source_answer": "2024=2^3×11×23；2025=3^4×5^2；999999=3^3×7×11×13×37；142857=3^3×11×13×37；101=101；1001=7×11×13；10001=73×137；10101=3×7×13×37；111=3×37；1111=11×101；11111=41×271；111111=3×7×11×13×37。", "source_solution": "原页以红笔逐项写出上述质因数分解，无额外文字推导。", "topics": [ "质因数分解", "唯一分解定理" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "无", "review": { "answer_status": "verified", "verified_answer": "2024=2^3×11×23；2025=3^4×5^2；999999=3^3×7×11×13×37；142857=3^3×11×13×37；101=101；1001=7×11×13；10001=73×137；10101=3×7×13×37；111=3×37；1111=11×101；11111=41×271；111111=3×7×11×13×37。", "solution_status": "consistent", "note": "逐项独立分解并检验均成立。原PDF第1页手写142857=3^3×11×13×37；题库旧转录误写为3×11×13×37，已更正source_answer。", "verified_solution": "先把一个数拆成小质数的乘积；直到每个因数都是质数，就完成了质因数分解。相同的质因数要写成幂，例如2×2×2写成2^3。逐项结果是：2024=2^3×11×23；2025=3^4×5^2；999999=3^3×7×11×13×37；142857=3^3×11×13×37；101是质数；1001=7×11×13；10001=73×137；10101=3×7×13×37；111=3×37；1111=11×101；11111=41×271；111111=3×7×11×13×37。每个等式右边相乘都等于左边，而且列出的每个因数本身都是质数。" }, "cross_reference_ids": [] }, { "id": "S03-Q002", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "练习1", "problem": "四个连续奇数的乘积是229425，那么这4个奇数的和是多少？", "source_answer": "88", "source_solution": "229425=3×5^2×7×19×23；原解析直接取四个连续奇数为19、21、23、25，和为19+21+23+25=88。", "topics": [ "质因数分解", "连续奇数", "乘积与和" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [ "题目只写“四个连续奇数”，没有说明是正奇数。若允许负奇数，-25、-23、-21、-19的乘积同样是229425，但和为-88；因此答案88依赖“正奇数”这一未写明的条件。" ], "duplicate_notes": "无", "review": { "answer_status": "conditional", "verified_answer": "若题意限定为四个连续正奇数，答案是88（19+21+23+25）。若允许负奇数，还可取-25、-23、-21、-19，和为-88；原题需补“正”才能唯一。", "solution_status": "incomplete", "note": "原卷第1页给出19、21、23、25并算得88；该序列确实满足乘积条件。题干未限定正数，故按整数解释不唯一。", "verified_solution": "先检查原卷给出的四个正奇数：19×21×23×25=229425，和是19+21+23+25=88。对正奇数来说，连续四项的乘积会随第一项变大而变大，所以这个正数解不会有第二组。可是题目没写“正奇数”：负数序列-25、-23、-21、-19的乘积也等于229425，和却是-88。为保证答案唯一，题干应改成“四个连续正奇数”。" }, "cross_reference_ids": [] }, { "id": "S03-Q003", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "例题2", "problem": "求下列各组数的最大公因数与最小公倍数：（1）288与360；60、96、144。（2）A=2×2×5与B=2×3×5；8!与2025。（3）（3569，2407）；（1273，1541）。", "source_answer": "（1）(288,360)=72，[288,360]=1440；(60,96,144)=12，[60,96,144]=1440。（2）(A,B)=10，[A,B]=60；(8!,2025)=45，[8!,2025]=1814400。（3）(3569,2407)=83；(1273,1541)=67。", "source_solution": "原解析用质因数分解及辗转相除法：8!=2^7×3^2×5×7，2025=3^4×5^2；3569−2407=1162，2407−1162=1245，1245−1162=83，1162÷83=14；1541÷1273余268，1273÷268余201，268÷201余67，201÷67=3。", "topics": [ "最大公因数与最小公倍数", "辗转相除法", "阶乘质因数分解" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "无", "review": { "answer_status": "verified", "verified_answer": "（1）288与360：最大公因数72，最小公倍数1440；60、96、144：最大公因数12，最小公倍数1440。（2）A=20、B=30：最大公因数10，最小公倍数60；8!与2025：最大公因数45，最小公倍数1814400。（3）(3569,2407)=83；(1273,1541)=67。", "solution_status": "incomplete", "note": "所有数值经质因数分解或辗转相除独立复算一致；原卷对多数组合只列结果，补入逐步说明。", "verified_solution": "两个数求最大公因数时，找它们共同拥有的质因数，并取较小的指数；求最小公倍数时，把出现过的质因数都取上，并取较大的指数。例如288=2^5×3^2、360=2^3×3^2×5，所以最大公因数是2^3×3^2=72，最小公倍数是2^5×3^2×5=1440。60、96、144的质因数分解分别为2^2×3×5、2^5×3、2^4×3^2，因此最大公因数为2^2×3=12，最小公倍数为2^5×3^2×5=1440。A=20、B=30，故最大公因数10、最小公倍数60。8!=40320=2^7×3^2×5×7，而2025=3^4×5^2，所以最大公因数为3^2×5=45，最小公倍数为2^7×3^4×5^2×7=1814400。最后两组用辗转相除：3569=2407+1162，2407=2×1162+83，1162=14×83，所以最大公因数是83；1541=1273+268，1273=4×268+201，268=201+67，201=3×67，所以最大公因数是67。" }, "cross_reference_ids": [] }, { "id": "S03-Q004", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "练习2", "problem": "（1）已知A、B两数的最小公倍数是360，最大公因数是6，若A=30，求B。（2）已知两个正整数的和是45，它们的最小公倍数是168，求这两个数。", "source_answer": "（1）B=72。（2）A=21，B=24（交换次序亦可）。", "source_solution": "（1）利用(A,B)×[A,B]=A×B，B=6×360÷30=72。（2）设A=ma、B=mb且(a,b)=1，则m(a+b)=45，mab=168；m为45与168的公因数，原解析分m=1、3，m=1时无解，m=3时a+b=15、ab=56，得a=7、b=8，故A=21、B=24。", "topics": [ "最大公因数与最小公倍数", "互质分解", "整数方程" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "无", "review": { "answer_status": "verified", "verified_answer": "（1）B=72。（2）这两个正整数是21和24（交换次序也可以）。", "solution_status": "incomplete", "note": "两问答案均正确；原解析称m=1时无解但未展示排除过程，补入易读的因数检查。", "verified_solution": "（1）两个数相乘，等于它们的最大公因数乘最小公倍数，所以30×B=6×360，B=72。回代可得最大公因数6、最小公倍数360。（2）设最大公因数是m。m既整除45（两数的和），也整除168（它们的最小公倍数），所以m只能是1或3。若m=1，两数乘积就等于168，但和为45：较小的数只能是1、2、3或4，对应较大的数为44、43、42或41，乘积分别是44、86、126、164，都不是168。若m=3，把两数写成3a和3b，则a+b=15，且最小公倍数是3ab=168，所以ab=56。56中和为15的一对因数是7和8，因此两数是21和24。" }, "cross_reference_ids": [] }, { "id": "S03-Q005", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "例题3", "problem": "2024有多少个因数？（1）有多少个偶因数？有多少个因数是92的倍数？（2）所有因数的和是多少？所有因数的倒数和是多少？（3）所有因数的乘积是多少？因数积的因数个数是多少？（4）以2024为分母的最简真分数有多少个？这些分数的和是多少？", "source_answer": "因数总数16；偶因数12；92的倍数因数4；因数和4320；因数倒数和540/253；因数积=2024^8=2^24×11^8×23^8；因数积的因数个数2025；最简真分数880个，和440。", "source_solution": "2024=2^3×11×23，τ(2024)=4×2×2=16。偶因数数目3×2×2=12；92=2^2×23，倍数因数数目2×2×1=4。σ(2024)=(1+2+4+8)(1+11)(1+23)=4320，倒数和=4320/2024=540/253。因数成对乘积为2024^8，因数积的质因数指数为24、8、8，故τ=25×9×9=2025。φ(2024)=2024(1−1/2)(1−1/11)(1−1/23)=880；最简真分数之和为φ(n)/2=440。", "topics": [ "因数个数", "因数和与倒数和", "欧拉函数", "因数积" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "无", "review": { "answer_status": "verified", "verified_answer": "2024有16个因数；偶因数12个；92的倍数因数4个；所有因数和4320；倒数和540/253；因数乘积2024^8=2^24×11^8×23^8；该乘积有2025个因数；以2024为分母的最简真分数有880个，分数和为440。", "solution_status": "consistent", "note": "全部结果独立复算：2024=2^3×11×23；按质因数指数计数、因数和公式及互质分数配对得到相同答案。", "verified_solution": "先分解2024=2^3×11×23。它的因数都可写成2^a×11^b×23^c，其中a可选0、1、2、3，b、c各可选0或1，所以共有4×2×2=16个。（1）偶因数要至少有一个2，指数有3种选法，故有3×2×2=12个；92=2^2×23，作为因数必须带有2^2或2^3，并且含23，11可选可不选，所以有2×2=4个。（2）因数和为(1+2+4+8)(1+11)(1+23)=4320。所有因数倒数相加等于因数和除以2024，因此是4320/2024=540/253。（3）把每个因数和它的配对因数相乘，得到2024；16个因数共有8对，所以乘积是2024^8=2^24×11^8×23^8。这个乘积的因数个数是(24+1)(8+1)(8+1)=2025。（4）分子与2024没有公因数的分数才是最简分数。这样的分子有2024×(1−1/2)(1−1/11)(1−1/23)=880个。每个分子a都能和2024−a配成一对，两份分数相加为1；880个分数分成440对，所以总和是440。" }, "cross_reference_ids": [] }, { "id": "S03-Q006", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "练习3", "problem": "240的因数有多少个？因数和是多少？所有因数的倒数之和等于多少？", "source_answer": "20个；744；31/10。", "source_solution": "240=2^4×3×5，τ=5×2×2=20；σ=(1+2+4+8+16)(1+3)(1+5)=744；倒数和=744/240=31/10。", "topics": [ "因数个数", "因数和与倒数和" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "无", "review": { "answer_status": "verified", "verified_answer": "240有20个因数；因数和为744；所有因数的倒数和为31/10。", "solution_status": "consistent", "note": "由240=2^4×3×5独立复算，三个结果均与原卷相同。", "verified_solution": "240=2^4×3×5。因数可以从2的0至4次方中选一种，再决定是否乘3、是否乘5，所以因数个数是5×2×2=20。因数和按各质因数分别相加再相乘：(1+2+4+8+16)(1+3)(1+5)=31×4×6=744。所有因数的倒数和等于744/240，约分后是31/10。" }, "cross_reference_ids": [] }, { "id": "S03-Q007", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "例题4", "problem": "把2028表示为若干个连续自然数（多于1个）的和，有____种不同的表示方法。", "source_answer": "5", "source_solution": "设为a+(a+1)+…+(a+n−1)=2028，则n(n+2a−1)=2×2028=2^3×3×13^2。原解析指出n与n+2a−1一奇一偶，计得5种。", "topics": [ "连续自然数求和", "奇偶性", "因数配对" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "无", "review": { "answer_status": "verified", "verified_answer": "5种：675+676+677；250+251+…+257；150+151+…+162；73+74+…+96；33+34+…+71。", "solution_status": "incomplete", "note": "答案5正确。原卷只写“2×3−1=5种”，没有列出奇因数如何对应连续和，已补出全部五种。", "verified_solution": "设首个自然数是a、共有n项，把等差数列求和式乘2，得 n(2a+n−1)=4056=2^3×3×13^2。两个因数n与2a+n−1一奇一偶，而且后者比n大。4056的奇因数是1、3、13、39、169、507；每个奇因数与4056相除得到一个因数对。奇因数1对应n=1，只有一项，不符合题意。其余因数对给出五组连续自然数：675—677、250—257、150—162、73—96、33—71。把每组首尾项求和再乘项数的一半，都得到2028，所以共有5种。" }, "cross_reference_ids": [] }, { "id": "S03-Q008", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "练习4", "problem": "2025最多可以表示成多少个连续自然数的和？", "source_answer": "54", "source_solution": "n(n+2a−1)=2025×2=2×3^4×5^2；取最大可行长度n=54，此时54+2a−1=75，a=11，故可表示为54个连续自然数。", "topics": [ "连续自然数求和", "因数分解", "最大长度" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "无", "review": { "answer_status": "verified", "verified_answer": "最多54个，具体为11+12+…+64=2025。", "solution_status": "incomplete", "note": "来源给出的54项构造正确，但原解没有证明不存在更长的表示，已补长度上界与候选检查。", "verified_solution": "设从a开始连续取n个自然数，则 n(2a+n−1)=2×2025=4050。由于a≥1，第二个因数比n大，所以n<√4050<64，即n≤63；同时n必须整除4050=2×3^4×5^2。55到63之间的数都不是4050的因数：55、57、59、61含有因数11、19或自身质数；56、58、62含有因数7、29或31；60需要两个因数2，但4050只有一个2；63含有因数7。因此比54大的项数都不行。54可行，因为4050÷54=75，故2a+54−1=75，得a=11。于是11+12+…+64共有54项，和为54×(11+64)÷2=2025。" }, "cross_reference_ids": [] }, { "id": "S03-Q009", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "例题5", "problem": "自然数N有30个正因数，求N的最小值。", "source_answer": "720", "source_solution": "τ(N)=30=30、15×2、10×3、6×5、5×3×2等指数分拆对应的候选：2^29，2^14×3，2^9×3^2，2^5×3^4，2^4×3^2×5=720；原解析取最小值720。", "topics": [ "因数个数", "最小值构造", "指数分拆" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "无", "review": { "answer_status": "verified", "verified_answer": "最小值为720。", "solution_status": "incomplete", "note": "30个因数对应的指数分组均已检查；原解列出候选但未说明为何指数应按降序配最小质数，已补最小性理由。", "verified_solution": "若N的质因数分解为 p1^e1×…×pk^ek，因数个数是(e1+1)…(ek+1)=30。30的乘法分组对应指数：29；14、1；9、2；5、4；4、2、1。要使N最小，先用最小的质数2、3、5，并把较大的指数给较小的质数；否则交换两个质数的指数会使乘积变小。各组最小候选依次为2^29、2^14×3、2^9×3^2、2^5×3^4、2^4×3^2×5=720。比较后720最小，且其因数个数为(4+1)(2+1)(1+1)=30。" }, "cross_reference_ids": [] }, { "id": "S03-Q010", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "练习5", "problem": "有8个因数且能被337整除的最小自然数是____。", "source_answer": "2022", "source_solution": "τ(N)=8=8、4×2、2×2×2。因337为质数，取N=2×3×337=2022，其因数个数为2×2×2=8；原解析比较结构后取2022。", "topics": [ "因数个数", "最小值构造", "质数整除" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "无", "review": { "answer_status": "verified", "verified_answer": "2022。", "solution_status": "incomplete", "note": "2022=2×3×337且因数数为8；已分类比较337的指数1、3、7与其余质因数分组，补出最小性。", "verified_solution": "337是质数。若N能被337整除且有8个因数，337的指数加1只能是8的因数2、4或8，所以337的指数只能是1、3或7。指数为1时，剩余部分的因数个数需为4：最小的“一个质数的三次方”是2^3=8；最小的“两个不同质数各一次”是2×3=6，所以更小的是6，得N=337×2×3=2022。指数为3时，剩余部分至少还要贡献2个因数，最小也要乘2，得到2×337^3，比2022大；指数为7时，N至少为337^7，更大。检查2022=2×3×337的因数个数为(1+1)^3=8，故最小值是2022。" }, "cross_reference_ids": [] }, { "id": "S03-Q011", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "例题6", "problem": "一个数有10个因数，它的11倍有12个因数，它的22倍有24个因数。那么，它的88倍有____个因数。", "source_answer": "48", "source_solution": "τ(N)=10且τ(11N)=12，排除N=a^9后取N=a×11^4，其中a为不同于11的质数。τ(22N)=24进一步排除a=2；于是88N=2^3×a×11^5，因数个数为(3+1)(1+1)(5+1)=48。", "topics": [ "因数个数", "指数变化", "分类讨论" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "无", "review": { "answer_status": "verified", "verified_answer": "48。", "solution_status": "incomplete", "note": "按因数个数10的两种指数结构分类后，条件唯一推出N=11^4×q且q不是2或11，结果48；原解省略了分类理由，已补全。", "verified_solution": "设N的质因数指数结构。因数个数10只可能来自N=p^9或N=p^4q（p、q为不同质数）。若N=p^9，11N的因数个数只能是11（p=11）或20（p≠11），不可能是12。若N=p^4q，11N有12个因数只在p=11时成立，于是N=11^4q，其中q是不同于11的质数。此时22N=2×11^5×q。若q=2，因数个数是(2+1)(5+1)=18，不是24，所以q不能是2；q也不能是11。因此88N=2^3×11^5×q，三个质因数指数分别为3、5、1，因数个数为(3+1)(5+1)(1+1)=48。" }, "cross_reference_ids": [] }, { "id": "S03-Q012", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "拓展（1）", "problem": "求11/49和2又9/56的最小公倍数和最大公因数。", "source_answer": "原解析写：最大公因数11/392，最小公倍数121/7。", "source_solution": "原页把混合数2又9/56化成121/56；按分数最大公因数=分子最大公因数/分母最小公倍数、分数最小公倍数=分子最小公倍数/分母最大公因数计算。", "topics": [ "分数的最大公因数与最小公倍数", "分数约分" ], "page_refs": { "problem": [ 5 ], "answer": [ 5 ], "solution": [ 5 ] }, "ambiguities": [ "原题第二个数是带分数2又9/56；题库旧题干漏掉整数部分2，现已按原页更正。原页手写将其化为121/56，题库旧解析转录将121误成12，现已修正。", "题目未先定义分数的最大公因数和最小公倍数；以下按原解析采用的常见竞赛约定核验。" ], "duplicate_notes": "无", "review": { "answer_status": "verified_with_caveat", "verified_answer": "按原卷采用的分数最大公因数/最小公倍数约定：最大公因数=11/392，最小公倍数=121/7。", "solution_status": "consistent", "note": "原PDF第5页印刷题面为11/49与混合数2又9/56，手写将后者化为121/56；题库旧转录漏了整数部分2并把121误写为12，均已纠正。按原卷所用分数公因数/公倍数约定，结果11/392与121/7正确；题干本身未解释该约定。", "verified_solution": "先把混合数化成假分数：2又9/56=(2×56+9)/56=121/56。按原卷的分数公因数和公倍数约定，分数最大公因数用“分子最大公因数÷分母最小公倍数”，分数最小公倍数用“分子最小公倍数÷分母最大公因数”。11和121的最大公因数是11、最小公倍数是121；49和56的最大公因数是7、最小公倍数是392。因此分数最大公因数为11/392，最小公倍数为121/7。" }, "cross_reference_ids": [] }, { "id": "S03-Q013", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "拓展（2）", "problem": "求2/15、4/21和8/35的最小公倍数和最大公因数。", "source_answer": "最大公因数2/105，最小公倍数8。", "source_solution": "原解析写：(2/15，4/21，8/35)=2/105；[2/15，4/21，8/35]=8。", "topics": [ "分数的最大公因数与最小公倍数" ], "page_refs": { "problem": [ 5 ], "answer": [ 5 ], "solution": [ 5 ] }, "ambiguities": [ "原题未定义分数的最大公因数和最小公倍数；按原卷采用的竞赛常见约定核验。" ], "duplicate_notes": "无", "review": { "answer_status": "verified_with_caveat", "verified_answer": "按原卷采用的分数公因数/公倍数约定：最大公因数=2/105，最小公倍数=8。", "solution_status": "incomplete", "note": "按原页所用约定独立复算，结果正确；原解析只列结果，已补计算。", "verified_solution": "三个分数均为最简分数。分数最大公因数按“分子最大公因数÷分母最小公倍数”计算：gcd(2,4,8)=2，lcm(15,21,35)=105，所以是2/105。分数最小公倍数按“分子最小公倍数÷分母最大公因数”计算：lcm(2,4,8)=8，gcd(15,21,35)=1，所以是8。原题未说明分数公因数/公倍数的定义，以上沿用原卷约定。" }, "cross_reference_ids": [] }, { "id": "S03-Q014", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "拓展（3）", "problem": "求3/28、15/56和1又1/20的最小公倍数和最大公因数。", "source_answer": "最大公因数3/280，最小公倍数105/4。", "source_solution": "原解析把1又1/20化为21/20，写：(3/28，15/56，21/20)=3/280；[3/28，15/56，21/20]=105/4。", "topics": [ "分数的最大公因数与最小公倍数", "带分数化假分数" ], "page_refs": { "problem": [ 5 ], "answer": [ 5 ], "solution": [ 5 ] }, "ambiguities": [ "原题未定义分数的最大公因数和最小公倍数；按原卷采用的竞赛常见约定核验。" ], "duplicate_notes": "无", "review": { "answer_status": "verified_with_caveat", "verified_answer": "按原卷采用的分数公因数/公倍数约定：最大公因数=3/280，最小公倍数=105/4。", "solution_status": "incomplete", "note": "将1又1/20化成21/20后按原页约定复算，结果正确；原解析只列结果，已补计算。", "verified_solution": "先把带分数化成假分数：1又1/20=(1×20+1)/20=21/20。分子3、15、21的最大公因数是3，分母28、56、20的最小公倍数是280，所以分数最大公因数为3/280。分子最小公倍数是105，分母最大公因数是4，所以分数最小公倍数为105/4。原题未说明分数公因数/公倍数的定义，以上沿用原卷约定。" }, "cross_reference_ids": [] }, { "id": "S03-Q015", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "作业1", "problem": "（1）四个连续偶数的乘积是5760，这四个数的平均数是____。（2）甲数=2×3×5，乙数=2×3×3，甲数和乙数的最大公因数是____，最小公倍数是____。", "source_answer": "（1）9；（2）最大公因数6，最小公倍数90。", "source_solution": "（1）5760=6×8×10×12，平均数=(6+8+10+12)÷4=9。（2）甲=2×3×5，乙=2×3^2，故GCD=2×3=6，LCM=2×3^2×5=90。", "topics": [ "连续偶数", "平均数", "最大公因数与最小公倍数" ], "page_refs": { "problem": [ 7 ], "answer": [ 7 ], "solution": [ 7 ] }, "ambiguities": [ "第（1）问只写“四个连续偶数”，未限定为正偶数；负数序列−12、−10、−8、−6也满足乘积条件，平均数为−9。" ], "duplicate_notes": "无", "review": { "answer_status": "conditional", "verified_answer": "（1）若为四个连续正偶数，平均数为9；若允许负偶数，还有−12、−10、−8、−6这一组，平均数为−9，因此原题按整数解释不唯一。（2）最大公因数6，最小公倍数90。", "solution_status": "incomplete", "note": "（1）原卷的正偶数列6、8、10、12乘积为5760，平均数9；但题干未限定正数，负数列−12、−10、−8、−6也满足、平均数−9。（2）两数30与18的最大公因数和最小公倍数为6和90。建议题干补“正偶数”。", "verified_solution": "（1）检查原卷列出的6、8、10、12：乘积为5760，平均数=(6+8+10+12)÷4=9。若四个数都要求为正，随着最小的偶数变大，四数乘积也会变大，所以这一组唯一。可是题干没有写“正偶数”；−12、−10、−8、−6的乘积同样是5760，平均数是−9。要让9成为唯一答案，题干应写“四个连续正偶数”。（2）甲数=2×3×5=30，乙数=2×3^2=18。共同质因数取较小指数得最大公因数2×3=6；全部质因数取较大指数得最小公倍数2×3^2×5=90。" }, "cross_reference_ids": [] }, { "id": "S03-Q016", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "作业2", "problem": "（1）2520的因数一共有多少个？有多少个偶因数？所有因数的倒数和是多少？（2）已知两个自然数的积为240，最小公倍数为60，求这两个数。", "source_answer": "（1）48个；36个；26/7。（2）4和60，或12和20。", "source_solution": "（1）2520=2^3×3^2×5×7，τ=4×3×2×2=48，偶因数=3×3×2×2=36；σ(2520)=9360，倒数和=9360/2520=26/7。（2）设A=ma、B=mb，LCM=mab=60且m^2ab=240，得m=4、ab=15，互质因子对为1×15或3×5，故(A,B)=(4,60)或(12,20)。", "topics": [ "因数个数", "偶因数", "因数倒数和", "最大公因数与最小公倍数", "整数分解" ], "page_refs": { "problem": [ 7 ], "answer": [ 7 ], "solution": [ 7 ] }, "ambiguities": [], "duplicate_notes": "无", "review": { "answer_status": "verified", "verified_answer": "（1）2520有48个因数，其中36个偶因数；所有因数倒数和为26/7。（2）这两个自然数是4和60，或12和20（交换顺序也可）。", "solution_status": "consistent", "note": "2520的质因数分解、因数个数、偶因数个数和倒数和均正确；乘积240且最小公倍数60的无序数对恰为(4,60)、(12,20)。", "verified_solution": "（1）2520=2^3×3^2×5×7。因数个数为(3+1)(2+1)(1+1)(1+1)=48。偶因数至少含一个2，指数有3种选法，所以有3×3×2×2=36个。所有因数和为(1+2+4+8)(1+3+9)(1+5)(1+7)=9360；因数倒数和等于9360÷2520=26/7。（2）设最大公因数为m，两数写成ma、mb，且a、b没有大于1的公因数。最小公倍数mab=60，乘积m^2ab=240，所以m=4、ab=15。15的互质因数对为1×15和3×5，得到(4,60)与(12,20)。代回两组乘积都是240，最小公倍数都是60。" }, "cross_reference_ids": [] }, { "id": "S03-Q017", "source_id": "S03", "source_filename": "26数论模块训练营第三讲解析(1).pdf", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "source_question_label": "作业3", "problem": "有16个因数的数最小是多少？", "source_answer": "120", "source_solution": "τ(N)=16的指数分拆对应候选：2^15，2^7×3，2^3×3^3，2^3×3×5，2×3×5×7；原解析列出候选并取N_min=120。", "topics": [ "因数个数", "最小值构造", "指数分拆" ], "page_refs": { "problem": [ 7 ], "answer": [ 7 ], "solution": [ 7 ] }, "ambiguities": [], "duplicate_notes": "与例题5同属最小值构造主题，但题目条件（16个因数）不同，不是重复题。", "review": { "answer_status": "verified", "verified_answer": "最小值为120。", "solution_status": "incomplete", "note": "16个因数对应的指数分组均已检查；原解析列出候选但未说明指数如何分配给最小质数，已补最小性理由。", "verified_solution": "因数个数公式要求质因数指数加1的乘积为16。指数分组只有15；7和1；3和3；3、1、1；1、1、1、1。要让数最小，使用最小质数2、3、5、7，并把较大指数给较小质数。各组最小候选为2^15=32768，2^7×3=384，2^3×3^3=216，2^3×3×5=120，2×3×5×7=210。最小的是120；它的因数个数为(3+1)(1+1)(1+1)=16。" }, "cross_reference_ids": [] }, { "id": "S04-Q001", "source_id": "S04", "source_filename": "26数论短期班小测1解析版(1).pdf", "lesson_or_test": "数论短期班 小测1", "source_question_label": "1", "problem": "若五位数12□38是9的倍数，则□内所填的数为____。", "source_answer": "4", "source_solution": "设方框中的数为a，由9的整除特征可知，9∣1 + 2 + a + 3 + 8，得a = 4。", "topics": [ "整除性与整除特征", "9的倍数判定" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "文档内未发现重复题。", "review": { "answer_status": "verified", "verified_answer": "4", "solution_status": "consistent", "note": "各位数字和为14+a；要被9整除需14+a=18，故a=4，且为数字，答案正确。", "verified_solution": "9的倍数有一个小窍门：各位数字相加也要是9的倍数。已知数字相加为1+2+3+8=14，所以方框里的数字要让14+a成为9的倍数。a只能是0到9，14+a在14到23之间；这个范围里只有18是9的倍数，因此a=4。" }, "cross_reference_ids": [] }, { "id": "S04-Q002", "source_id": "S04", "source_filename": "26数论短期班小测1解析版(1).pdf", "lesson_or_test": "数论短期班 小测1", "source_question_label": "2", "problem": "四位数a72b能被36整除，这个四位数最大是____。", "source_answer": "9720", "source_solution": "", "topics": [ "整除性与整除特征", "36的倍数判定", "最值问题" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [ "第1页第2题的【解析】标题下为空，来源解析缺失。" ], "duplicate_notes": "文档内未发现重复题。", "review": { "answer_status": "verified", "verified_answer": "9720。", "solution_status": "not_provided", "note": "36=4×9。为使a72b最大先取a=9；9的倍数条件给出18+b为9的倍数，b∈{0,9}；末两位为20或29时仅20能被4整除，因此得9720，且9720÷36=270。原文有解析标题但未提供解析内容。", "verified_solution": "要使四位数最大，先取千位 a=9。能被 9 整除要求 9+7+2+b=18+b 为 9 的倍数，所以 b=0 或 9。末两位 20 能被 4 整除，29 不能，因此 b=0，最大数为 9720；9720÷36=270。" }, "cross_reference_ids": [] }, { "id": "S04-Q003", "source_id": "S04", "source_filename": "26数论短期班小测1解析版(1).pdf", "lesson_or_test": "数论短期班 小测1", "source_question_label": "3", "problem": "四位数17□□能被53整除，这个四位数是____。", "source_answer": "1749", "source_solution": "用1799试除，得33余50，所以这个四位数为1799 − 50 = 1749。", "topics": [ "整除性与整除判定", "带余除法", "整除范围搜索" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "文档内未发现重复题。", "review": { "answer_status": "verified", "verified_answer": "1749", "solution_status": "consistent", "note": "1799=53×33+50，减去余数得1749=53×33；在1700至1799范围内，53的倍数只有53×33=1749，答案正确。", "verified_solution": "四位数17□□在1700到1799之间。53×32=1696，比1700小；53×33=1749，在范围内；53×34=1802，比1799大。因此唯一的四位数是1749。" }, "cross_reference_ids": [] }, { "id": "S04-Q004", "source_id": "S04", "source_filename": "26数论短期班小测1解析版(1).pdf", "lesson_or_test": "数论短期班 小测1", "source_question_label": "4", "problem": "正整数 n 使得 2n−1 整除 20n，求 n 的值。", "source_answer": "1，3", "source_solution": "", "topics": [ "整除性", "最大公因数", "因数枚举" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [] }, "ambiguities": [ "第1页第4题未标出【解析】段，仅有题面和“答案：1，3”。" ], "duplicate_notes": "文档内未发现重复题。", "review": { "answer_status": "verified", "verified_answer": "n=1 或 n=3。", "solution_status": "not_provided", "note": "令d=2n−1。因gcd(2n−1,n)=1，且d∣20n，所以d∣20；d为正奇数，故d∈{1,5}，分别得n=1,3，答案正确。逐一代入亦成立。原文仅给出答案，未给出解析。", "verified_solution": "令 d=2n−1。因 gcd(2n−1,n)=1，且 d |20n，所以 d |20。d 是正奇数，只能为 1 或 5；由 n=(d+1)/2，得 n=1 或 3。代回原式都成立。" }, "cross_reference_ids": [] }, { "id": "S05-Q001", "source_id": "S05", "source_filename": "26数论模块班小测2解析版(1).pdf", "lesson_or_test": "数论模块班小测2", "source_question_label": "1", "problem": "（2分）如果两位数好玩和玩好都是质数，那么，这两个质数的和的最大值是____。", "source_answer": "176", "source_solution": "97 + 79 = 176．", "topics": [ "质数与合数", "数字倒置与最值" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "176", "solution_status": "incomplete", "note": "来源解析只列出97+79=176，未证明最大性。本轮独立复核：数字和18只能取9+9，得到99不是质数；数字和17只能取8和9，倒置数98是大于2的偶数，不是质数；数字和16可由79和97实现且两者都是质数。", "verified_solution": "设两个倒置数的数字为 x、y，则它们的和为 11(x+y)。若 x+y=18，只能是9+9，得到99，不是质数；若 x+y=17，只能是8和9，倒置数中98是大于2的偶数，也不是质数。因此 x+y≤16。取7和9，得到79、97，两者都是质数，和为176，所以最大值是176。" }, "cross_reference_ids": [] }, { "id": "S05-Q002", "source_id": "S05", "source_filename": "26数论模块班小测2解析版(1).pdf", "lesson_or_test": "数论模块班小测2", "source_question_label": "2", "problem": "（3分）200到220之间有唯一的质数，它是____。", "source_answer": "211", "source_solution": "范围内的偶数全为合数；201、207、213、219都是3的倍数；205、215是5的倍数；203、217是7的倍数；209是11的倍数；只有211是质数。", "topics": [ "质数与合数", "整除与倍数" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "211", "solution_status": "consistent", "note": "200—220内除211外的奇数均可由题解所列小质因数整除；211不被不超过√211的质数2、3、5、7、11、13整除，确为质数。", "verified_solution": "200到220之间的偶数都大于2，所以不是质数。奇数候选除211外都能被小质数整除：201、207、213、219能被3整除；203、217能被7整除；205、215能被5整除；209能被11整除。检查211：它不被2、3、5、7、11或13整除，而这些是小于√211的所有质数，所以211是质数，也是区间里唯一的质数。" }, "cross_reference_ids": [] }, { "id": "S05-Q003", "source_id": "S05", "source_filename": "26数论模块班小测2解析版(1).pdf", "lesson_or_test": "数论模块班小测2", "source_question_label": "3", "problem": "（3分）2154年，潘多拉星球的纳美族公主在神树中发现一个神秘方程，这个记录于2035年的方程隐藏了阿凡达计划的重要信息，这个方程是：a^a·b·c + b = 2035，已知a、b、c都是质数，则a + b + c =____。", "source_answer": "42", "source_solution": "由于a^a·b·c + b = 2035 = 5×11×37，可知a^a·b·c与b一定为一奇一偶，则偶质数2一定为a或c。若a=2，则4bc+b=b·(4c+1)=5×11×37，此时无论b为5、11或37，均没有质数c可对应；则c=2，故2a^a b+b=b·(2a^a+1)=5×11×37，此时，当b=37时，a=3，∴a+b+c=3+37+2=42。", "topics": [ "质数与合数", "奇偶性", "因式分解与整除" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [ "题面幂与乘法的排版较紧，按页面可辨内容解释为a^a·b·c+b=2035；若原意不是a的a次方，结论可能改变。" ], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "42", "solution_status": "consistent", "note": "b不能为2（左式为偶数而2035为奇数），故b为奇质数；由奇偶性a或c为2。a=2时b=5、11、37分别不给出质数c；c=2时b=37给出2a^a+1=55，即a^a=27、a=3，故和为42。", "verified_solution": "2035=5×11×37，而且是奇数。b不能是2，否则左边两项都是偶数；因此b是奇质数。为了两项相加得到奇数2035，a或c必须是偶质数2。若a=2，方程变成b(4c+1)=2035；把b取5、11、37逐一代入，都不能得到质数c。于是c=2，方程变为b(2a^a+1)=2035。b只能是5、11、37；前两种会要求a^a分别等于203或92，都不是质数的自身幂。b=37时得到a^a=27，所以a=3。三数是3、37、2，和为42。" }, "cross_reference_ids": [] }, { "id": "S05-Q004", "source_id": "S05", "source_filename": "26数论模块班小测2解析版(1).pdf", "lesson_or_test": "数论模块班小测2", "source_question_label": "4", "problem": "（3分）有两个两位数，它们的乘积是2132，如果它们的和是奇数，那么它们的和等于____。", "source_answer": "93", "source_solution": "因为奇数+偶数=奇数。所以当它们的和是奇数时，这两个两位数一个是奇数一个是偶数，2132=2×2×13×41=52×41，所以这两个数分别是52和41，那这两个数的和是41+52=93。", "topics": [ "奇偶性", "因数与因数分解", "两位数" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "93", "solution_status": "consistent", "note": "2132的两位数因数对中，满足一奇一偶的为41和52；其和为93。其他因数对不同时为两位数或为偶偶。", "verified_solution": "2132=2²×13×41。把因数两两配成乘积2132，且都要是两位数，可能的因数对只有26×82和41×52。第一组都是偶数，和是偶数；第二组一奇一偶，和是奇数。因此所求和是41+52=93。" }, "cross_reference_ids": [] }, { "id": "S05-Q005", "source_id": "S05", "source_filename": "26数论模块班小测2解析版(1).pdf", "lesson_or_test": "数论模块班小测2", "source_question_label": "5", "problem": "（3分）A、B两个自然数的最大公因数是12，A×B=1728，A、B的最小公倍数是____。", "source_answer": "144", "source_solution": "1728÷12=144，所以A、B的最小公倍数为144。", "topics": [ "最大公因数与最小公倍数", "因数关系" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "144", "solution_status": "consistent", "note": "对自然数A、B恒有AB=gcd(A,B)×lcm(A,B)，故lcm=1728÷12=144。", "verified_solution": "两个数相乘，等于它们的最大公因数乘最小公倍数：A×B=gcd(A,B)×lcm(A,B)。所以lcm(A,B)=1728÷12=144。" }, "cross_reference_ids": [] }, { "id": "S05-Q006", "source_id": "S05", "source_filename": "26数论模块班小测2解析版(1).pdf", "lesson_or_test": "数论模块班小测2", "source_question_label": "6", "problem": "（3分）如图是A、B、C三个互相咬合的齿轮，若A轮转3圈，B轮转7圈，C轮转2圈，那么这三个齿轮的齿数最少的是A轮____齿，B轮____齿，C轮____齿。", "source_answer": "14 6 21", "source_solution": "A轮、B轮及C轮转过的总齿数是相等的，并且3、7、2三个数两两互质，所以[7,2,3]=42，A轮齿数为42÷3=14；B轮齿数为42÷7=6；C轮齿数为42÷2=21。由题意可知A轮转3圈、B轮转7圈、C轮转2圈，记A、B、C三个齿轮的周长分别是a、b、c，则3a=7b=2c，可得a:b:c=14:6:21，因此三个齿轮的齿数最少分别是14齿、6齿、21齿。故答案为：14；6；21。", "topics": [ "最小公倍数", "比例", "齿轮问题" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "A=14，B=6，C=21", "solution_status": "consistent", "note": "啮合齿轮每圈经过的齿数相同，故3A=7B=2C；取公共值lcm(3,7,2)=42得A=14、B=6、C=21，且三者不能同时再缩小。", "verified_solution": "齿轮咬合时，接触处经过的齿数相同。A转3圈、B转7圈、C转2圈，因此3A=7B=2C。这个共同的齿数必须同时是3、7、2的倍数，最小是它们的最小公倍数42。于是A=42÷3=14，B=42÷7=6，C=42÷2=21。" }, "cross_reference_ids": [] }, { "id": "S05-Q007", "source_id": "S05", "source_filename": "26数论模块班小测2解析版(1).pdf", "lesson_or_test": "数论模块班小测2", "source_question_label": "7", "problem": "（3分）已知a=2×2×3×5，b=2×3×m，如果a和b的最大公因数是12，那么m=____，a和b的最小公倍数是____。", "source_answer": "2；60", "source_solution": "先用短除法结合a，b最大公因数为12，可求出m=2，再继续结合短除法可得到a，b分别除以最大公因数后为5和1，所以[a,b]=12×5×1=60。由于a，b最大公因数为12，2×3=6，12÷6=2，则m是2的倍数，而m还是一个质数，所以m=2，a，b最小公倍数为12×5×1=60。", "topics": [ "最大公因数与最小公倍数", "质因数分解", "整除条件" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [ "第2页题面没有看到“m为质数”或等价条件；解析末尾自行称m为质数，属于题面缺失条件。" ], "duplicate_notes": "无重复题。", "review": { "answer_status": "conditional", "verified_answer": "按正整数 m 理解：gcd(10,m)=2，最小公倍数=30m，解不唯一。只有另补“m 为质数”时，才有 m=2、最小公倍数 60。", "solution_status": "error", "note": "题面没有“m为质数”条件。来源解答自行补入该条件；例如m=4时a=60、b=24，最大公因数仍为12，但最小公倍数为120。 本轮独立复核（原PDF第2页，2026-10-08）：目视确认题干未出现“m为质数”；独立验算m=4、6等也满足最大公因数为12，故来源的m=2、最小公倍数60仅在补充“m为质数”时成立。", "verified_solution": "a=60，b=6m。gcd(a,b)=gcd(60,6m)=6·gcd(10,m)=12，故 gcd(10,m)=2；这并不推出 m=2，例如 m=4、6、8 都满足。最小公倍数为 ab/gcd(a,b)=60×6m/12=30m。若原题漏印“m 为质数”，才可进一步得 m=2、最小公倍数 60。" }, "cross_reference_ids": [] }, { "id": "S06-Q001", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "例题1（1）", "problem": "已知n是正整数，则下列数中一定能整除(2n+3)^2−25的是（A.6 B.3 C.4 D.5）。", "source_answer": "C（4）", "source_solution": "(2n+3+5)(2n+3−5)=(2n+8)(2n−2)=4(n+4)(n−1)，所以一定被4整除。", "topics": [ "整除性与因式分解" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "C（4）。", "solution_status": "consistent", "note": "恒等变形为4(n+4)(n−1)，故任意正整数n都被4整除；其他选项可用n=2或3排除。", "verified_solution": "(2n+3)^2−25=(2n+3−5)(2n+3+5)=(2n−2)(2n+8)=4(n−1)(n+4)，因此总能被4整除。要排除其他选项，取n=3时原式为56，不能被3或6整除；取n=2时原式为24，不能被5整除。所以只有4一定能整除。" }, "cross_reference_ids": [] }, { "id": "S06-Q002", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "例题1（2）", "problem": "2^48−1能被60到70之间的某两个整数整除，则这两个数是（A.61和63 B.63和65 C.65和67 D.64和67）。", "source_answer": "B（63和65）", "source_solution": "2^48−1=(2^24+1)(2^12+1)(2^6+1)(2^3+1)(2^3−1)，其中65=2^6+1、9=2^3+1、7=2^3−1；并且63=2^6−1也整除2^48−1。", "topics": [ "幂的整除性", "因式分解" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "B（63和65）。", "solution_status": "consistent", "note": "因式分解与原式相等；63、65均为60到70间的因数，选项其余项分别被奇偶性或模运算排除。", "verified_solution": "反复使用平方差公式：2^48−1=(2^24−1)(2^24+1)=(2^12−1)(2^12+1)(2^24+1)=(2^6−1)(2^6+1)(2^12+1)(2^24+1)。其中2^6−1=63、2^6+1=65，所以两数都整除原数，并都在60到70之间。选项A中的61余33，选项C中的67余61；选项D含偶数64，而2^48−1是奇数。因此只有B正确。" }, "cross_reference_ids": [] }, { "id": "S06-Q003", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "练习1", "problem": "设a，b，c，d为正整数，且a^7=b^6，c^3=d^2，c−a=17，则d−b=____。", "source_answer": "601（手写计算为729−128=601）", "source_solution": "令b/a=m，则a=m^6；令d/c=n，则c=n^2。于是n^2−m^6=(n−m^3)(n+m^3)=17，得n=9、m^3=8、m=2，从而a=64，b=128，c=81，d=729，d−b=601。", "topics": [ "幂的整数结构", "因式分解" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "601。", "solution_status": "consistent", "note": "由质数指数比较，a=m^6、b=m^7、c=n^2、d=n^3；因数分解得到唯一正整数解m=2、n=9。", "verified_solution": "若a^7=b^6，比较两边每个质因数的指数，因为6与7互质，可写成a=m^6、b=m^7。类似地，c^3=d^2可写成c=n^2、d=n^3。代入c−a=17，得到(n−m^3)(n+m^3)=17。17是质数，两个正因数只能是1和17，所以n−m^3=1、n+m^3=17，解得n=9、m^3=8、m=2。于是b=128、d=729，d−b=601。" }, "cross_reference_ids": [] }, { "id": "S06-Q004", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "例题2（1）", "problem": "若x、y为正整数，xy−5x+6y=78，求(x,y)的解有多少组？", "source_answer": "5组", "source_solution": "x(y−5)+6(y−5)=48，故(x+6)(y−5)=48。正因数配对给出5组。", "topics": [ "二元一次丢番图方程", "因式分解计数" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "5组：(2,11)、(6,9)、(10,8)、(18,7)、(42,6)。", "solution_status": "consistent", "note": "变形为正整数因数对；要求x+6≥7，48的正因数对恰给出5组。", "verified_solution": "把原式移项并分组：xy−5x+6y−78=0，等价于(x+6)(y−5)=48。因为x为正整数，x+6≥7；等式右边为正，所以y−5也为正。48的因数对中，满足第一因数至少为7的有(8,6)、(12,4)、(16,3)、(24,2)、(48,1)，对应(x,y)分别为(2,11)、(6,9)、(10,8)、(18,7)、(42,6)，共5组。" }, "cross_reference_ids": [] }, { "id": "S06-Q005", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "例题2（2）", "problem": "下列方程各有多少组正整数解？(a) 3xy−x+6y−42=0；(b) 2x^2y−xy=8x+84。", "source_answer": "(a)3组；(b)2组", "source_solution": "(a) 3y(x+2)−(x+2)=40，即(x+2)(3y−1)=40，列出3组。(b) 原手写解析写成xy(2x−1)−4(2x−1)=88，得(x,y)=(1,92)、(6,2)，共2组。", "topics": [ "二元一次丢番图方程", "二元高次丢番图方程", "因式分解" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "(a) 3组：(3,3)、(6,2)、(18,1)；(b) 2组：(1,92)、(6,2)。", "solution_status": "consistent", "note": "回看原PDF第2页，(xy−4)(2x−1)=88展开后正好等价于原方程，先前“因式分解不等价”的复核结论是误判，已撤销；两部分答案穷尽后均正确。", "verified_solution": "(a) 原式等价于(x+2)(3y−1)=40。令u=x+2、v=3y−1，则u≥3，v为正且v除以3余2。40的有序因数对中符合条件的是(5,8)、(8,5)、(20,2)，得到(3,3)、(6,2)、(18,1)，共3组。(b) 原方程等价于xy(2x−1)=8x+84；两边同减4(2x−1)，得到(xy−4)(2x−1)=88。因为x、y为正整数，两个因数都为正数；2x−1是88的奇因数，只能是1或11。取1得x=1、y=92；取11得x=6、y=2。故恰有2组。" }, "cross_reference_ids": [] }, { "id": "S06-Q006", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "练习2（1）", "problem": "已知x、y为正整数，而xy−3x+8y=48，求(x,y)的解组数目。", "source_answer": "2组", "source_solution": "x(y−3)+8(y−3)=24，(x+8)(y−3)=24，因数配对得到2组。", "topics": [ "二元一次丢番图方程", "因式分解计数" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "2组：(4,5)、(16,4)。", "solution_status": "consistent", "note": "等价于(x+8)(y−3)=24；正整数约束下只允许因数对(12,2)、(24,1)，代回均成立。", "verified_solution": "把xy−3x+8y=48移项整理：x(y−3)+8(y−3)=24，所以(x+8)(y−3)=24。因x>0，x+8≥9；因乘积为正，y−3>0。24中第一因数至少为9的正因数对只有(12,2)和(24,1)，对应(x,y)=(4,5)和(16,4)，共2组。代回原式均成立。" }, "cross_reference_ids": [] }, { "id": "S06-Q007", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "练习2（2）", "problem": "整数a，b满足6ab=9a−10b+303，则a+b=____。", "source_answer": "15", "source_solution": "(3a+5)(2b−3)=288=2^5×3^2，取3a+5=32、2b−3=9，得a=9、b=6，所以a+b=15。", "topics": [ "二元丢番图方程", "因式分解" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "a=9、b=6，a+b=15。", "solution_status": "incomplete", "note": "将原方程化为(3a+5)(2b−3)=288。须同时使用2b−3为奇数和3a+5≡2(mod 3)筛选带符号因数；原解析只写出一组未证明唯一。", "verified_solution": "令u=3a+5、v=2b−3。原式等价于uv=288。因为v是奇数，288=2^5×9中的全部2因子都在u里，所以u只能是±32、±96、±288。又u=3a+5，故u除以3余2；上述数中只有u=32符合。于是v=288/32=9，解得a=9、b=6，和为15。" }, "cross_reference_ids": [] }, { "id": "S06-Q008", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "例题3", "problem": "已知等式1/15=1/a+1/b，其中a、b是非零自然数，求a+b的最大值。", "source_answer": "256", "source_solution": "ab−15a−15b=0，(a−15)(b−15)=15^2=3^2×5^2。取正因数配对1与225，得a=16、b=240，最大和为256。", "topics": [ "埃尔德什-施特劳斯型分数方程", "因式分解与因数配对" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [ "题面写“非零自然数”，通常自然数已为正，但若所在教材允许自然数含0，因分母不能为0，实际仍不影响可行解。" ], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "最大值为256，在(a,b)=(16,240)或(240,16)时取得。", "solution_status": "incomplete", "note": "变形正确。原解析指出因数对1与225，但未说明最大和理由；已用正因数和的不等式补足。", "verified_solution": "由1/15=1/a+1/b，通分得ab=15a+15b，即(a−15)(b−15)=225。因为a、b是正整数，且两个正倒数之和为1/15，所以a、b都大于15。令r=a−15、s=b−15，则r、s为正整数且rs=225。由(r−1)(s−1)≥0可得r+s≤rs+1=226，因此a+b=30+r+s≤256。取r=1、s=225（或交换），即a=16、b=240，确实达到256。" }, "cross_reference_ids": [ "XR-05" ] }, { "id": "S06-Q009", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "练习3", "problem": "在等式1/8=1/a+1/b中，a、b表示不同的自然数，要使等式成立，求a，b分别是多少？", "source_answer": "(a,b)=(9,72)、(10,40)、(12,24)（及交换次序）", "source_solution": "ab−8a−8b=0，(a−8)(b−8)=64=2^6。取正因数1、2、4及对应64、32、16，得(a,b)=(9,72)、(10,40)、(12,24)。", "topics": [ "分数方程", "因式分解与因数配对" ], "page_refs": { "problem": [ 2, 3 ], "answer": [ 2, 3 ], "solution": [ 2, 3 ] }, "ambiguities": [], "duplicate_notes": "第3页再次完整呈现与第2页末尾相同的练习3，标记为文档重复/跨页重复，仅计1道独立题。", "review": { "answer_status": "verified", "verified_answer": "无序解为(9,72)、(10,40)、(12,24)；每组交换次序也成立。", "solution_status": "consistent", "note": "由因数分解得到三组不同自然数，交换顺序后共6个有序解；原页列出的数对正确。", "verified_solution": "由1/8=1/a+1/b得ab−8a−8b=0，即(a−8)(b−8)=64。因为两个正倒数之和为1/8，a、b都大于8，所以两个因数为正。64的正因数配对为1×64、2×32、4×16、8×8；最后一对会得到a=b=16，违反“不同”。其余三对给出无序解(9,72)、(10,40)、(12,24)，各自交换a、b也成立。" }, "cross_reference_ids": [ "XR-03" ] }, { "id": "S06-Q010", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "例题4", "problem": "求1/x−1/y=1/2025有多少组不同的正整数解。", "source_answer": "22组", "source_solution": "2025y−2025x=xy，整理为(2025+y)(2025−x)=2025^2=3^8×5^4。2025^2的正因数个数为(8+1)(4+1)=45，因x、y正整数且排除中间情形，配对数为(45−1)/2=22组。", "topics": [ "分数方程", "因式分解与因数个数" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "22组不同的正整数解。", "solution_status": "incomplete", "note": "因数计数45及答案22正确；原解析未充分说明正解与小于2025的因数一一对应，已补出双向对应。", "verified_solution": "整理方程可得(2025+y)(2025−x)=2025^2。由于x、y为正整数，因数2025−x必须是小于2025的正因数；反过来，每个小于2025的正因数d都唯一给出x=2025−d、y=2025^2/d−2025，且二者为正整数。因此解数等于2025^2的正因数中小于平方根2025的个数。2025^2=3^8×5^4，共有(8+1)(4+1)=45个因数；其中1个等于2025，其余44个两两成互补因数对，故小于2025的有22个，解也有22组。" }, "cross_reference_ids": [] }, { "id": "S06-Q011", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "例题5", "problem": "求方程1/x+1/y+1/z=5/6的正整数解。", "source_answer": "15组", "source_solution": "不妨令x≥y≥z≥2，则1/x≤1/y≤1/z。由1/z≥5/18且z≤18/5，得z=2或3。z=3时1/x+1/y=1/2，(x−2)(y−2)=4，得(6,3,3)、(4,4,3)；z=2时1/x+1/y=1/3，(x−3)(y−3)=9，得(12,4,2)、(6,6,2)。按排列计数共3+3+6+3=15组。", "topics": [ "单位分数方程", "因式分解与分类枚举" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "15组正整数有序解。", "solution_status": "incomplete", "note": "总数15正确；原手写在z=2、3的因数配对处把x−常数与y−常数的标签顺序写反，但最终列出的三元组正确，现按x≥y≥z重排并说明排列数。", "verified_solution": "不妨设x≥y≥z。若z=1，倒数和已大于5/6，不可能；又5/6≤3/z，所以z≤18/5，故z只能是2或3。z=3时，1/x+1/y=1/2，化为(x−2)(y−2)=4；按x≥y，得到(x,y)=(6,3)或(4,4)，对应三元组(6,3,3)有3种排列、(4,4,3)有3种。z=2时，1/x+1/y=1/3，化为(x−3)(y−3)=9；按x≥y，得到(12,4)或(6,6)，对应(12,4,2)有6种排列、(6,6,2)有3种。合计3+3+6+3=15组。" }, "cross_reference_ids": [] }, { "id": "S06-Q012", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "例题6", "problem": "已知a是自然数，问a^4−3a^2+9是质数还是合数？给出你的证明。", "source_answer": "a=1时质数7；a=2时质数13；a≥3时合数", "source_solution": "a^4−3a^2+9=(a^2+3)^2−(3a)^2=(a^2−3a+3)(a^2+3a+3)。当a^2−3a+3=1时(a−1)(a−2)=0，即a=1或2，分别得7、13为质数；a≥3时两因子均大于1，为合数。", "topics": [ "质数与合数", "平方差因式分解" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [ "“自然数”是否包含0存在教材约定差异；若包含0，原解析需补a=0，结果为9（合数）。" ], "duplicate_notes": "", "review": { "answer_status": "verified_with_caveat", "verified_answer": "当a=1、2时分别为质数7、13；其他允许的自然数取值都得到合数。", "solution_status": "incomplete", "note": "因式分解正确；原解析按正自然数讨论，若“自然数”含0，须补a=0时数值9为合数。补充后结论完整。", "verified_solution": "a^4−3a^2+9=(a^2+3)^2−(3a)^2=(a^2−3a+3)(a^2+3a+3)。a=1时结果7，是质数；a=2时结果13，是质数。a≥3时两个因数都大于1，所以结果是合数。若教材把0也算作自然数，则a=0时结果为9，也是合数；因此无论采用哪种自然数约定，只有a=1、2时结果为质数。" }, "cross_reference_ids": [] }, { "id": "S06-Q013", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "拓展", "problem": "已知两个自然数的积与和之差恰等于它们的最大公约数与最小公倍数之和，求这样的自然数。", "source_answer": "(4,6)、(3,6)、(4,4)（交换次序视为同组）", "source_solution": "设两数A=ma、B=mb，且(a,b)=1，则AB−(A+B)=m+ m a b，化为(m−1)ab=1+a+b，即m−1=1/(ab)+1/a+1/b≤3，故m=2、3、4。分别得到m=2时(a,b)=(2,3)，A,B=(4,6)；m=3时(a,b)=(1,2)，A,B=(3,6)；m=4时(a,b)=(1,1)，A,B=(4,4)。", "topics": [ "最大公约数与最小公倍数", "互质分解", "整除性" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [ "题面排版中‘积与和之差’按AB−(A+B)理解；若将‘和之差’作其他括号解释则语义不明确。" ], "duplicate_notes": "", "review": { "answer_status": "verified_with_caveat", "verified_answer": "按原页通常解释AB−(A+B)：无序数对为(4,6)、(3,6)、(4,4)；若区分A、B，前两组各可交换顺序。", "solution_status": "incomplete", "note": "三组代入AB−(A+B)=gcd(A,B)+lcm(A,B)均成立；原题“积与和之差”按AB−(A+B)理解，已在答案和解释中明示。", "verified_solution": "设A=ma、B=mb，其中m=gcd(A,B)，且a、b互质，则lcm(A,B)=mab。按原题通常读法，AB−(A+B)=gcd+lcm，化为(m−1)ab=1+a+b，即m−1=1/a+1/b+1/(ab)≤3，所以m只能是2、3、4。m=2时，(a−1)(b−1)=2，得(a,b)=(2,3)或(3,2)，于是{A,B}={4,6}。m=3时，(2a−1)(2b−1)=3，得(a,b)=(1,2)或(2,1)，于是{A,B}={3,6}。m=4时，(3a−1)(3b−1)=4；两因子均除以3余2，故只能都为2，得a=b=1、A=B=4。" }, "cross_reference_ids": [] }, { "id": "S06-Q014", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "作业1", "problem": "若x、y为正整数，xy−3x+7y=51，求(x,y)的解组数目。", "source_answer": "3组", "source_solution": "x(y−3)+7(y−3)=51−21=30，(x+7)(y−3)=30。正因数配对给出3组。", "topics": [ "二元一次丢番图方程", "因式分解计数" ], "page_refs": { "problem": [ 6 ], "answer": [ 6 ], "solution": [ 6 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "3组：(3,6)、(8,5)、(23,4)。", "solution_status": "incomplete", "note": "变形正确；正整数限制下x+7≥8，只有三个可行正因数对，答案3组通过。", "verified_solution": "把方程整理为(x+7)(y−3)=30。因为x为正整数，x+7至少为8；乘积为正，所以y−3也为正。30的正因数对中，第一因数不小于8的只有(10,3)、(15,2)、(30,1)，分别得到(x,y)=(3,6)、(8,5)、(23,4)，共3组。" }, "cross_reference_ids": [] }, { "id": "S06-Q015", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "作业2", "problem": "方程1/x+1/y=1/77中的x、y均取正整数，得到的解(x,y)叫做方程的一个正整数解，则这个方程的正整数解有____个。", "source_answer": "9个", "source_solution": "xy−77x−77y=0，(x−77)(y−77)=77^2=7^2×11^2，τ(7^2×11^2)=3×3=9。", "topics": [ "分数方程", "因式分解与因数个数" ], "page_refs": { "problem": [ 6 ], "answer": [ 6 ], "solution": [ 6 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "9个有序正整数解。", "solution_status": "incomplete", "note": "变形及因数个数9正确；已说明每个正因数对应一个有序解，且x=y的一组也计入。", "verified_solution": "由1/x+1/y=1/77得xy−77x−77y=0，即(x−77)(y−77)=77^2=7^2×11^2。因x、y是正整数，原式每个倒数都小于1/77，所以x、y均大于77；两个因数都是正整数。任取77²的正因数d，就唯一得到有序解x=77+d、y=77+77²/d。因此解数等于77²的正因数个数：(2+1)(2+1)=9。包括x=y=154这一组。" }, "cross_reference_ids": [] }, { "id": "S06-Q016", "source_id": "S06", "source_filename": "26数论模块训练营第四讲解析(1).pdf", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "source_question_label": "作业3", "problem": "若n为正整数，且2^4+2^7+2^n是完全平方数，求正整数n的值。", "source_answer": "n=8", "source_solution": "2^4+2^7+2^n=144+2^n=m^2，故(m−12)(m+12)=2^n。两因子相差24且均为2的幂，只能为8、32，得m=20、2^n=256=2^8，故n=8。", "topics": [ "完全平方数", "幂的分解与奇偶性" ], "page_refs": { "problem": [ 6 ], "answer": [ 6 ], "solution": [ 6 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "n=8。", "solution_status": "incomplete", "note": "平方差因式分解正确；原解没有说明为何仅有8、32这组2的幂相差24，已补唯一性证明。", "verified_solution": "令完全平方数为m²。因为2^4+2^7=144=12²，故(m−12)(m+12)=2^n。m>12，所以两个因数都是正整数；乘积是2的幂，两个因数本身也必须是2的幂。设较小因数为2^r、较大为2^s（s>r），它们相差24：2^r(2^(s−r)−1)=24=8×3。括号内是奇数，因此2^r只能是8，括号为3，从而s−r=2，因数恰为8和32。于是m=20，2^n=8×32=256=2^8，得n=8。" }, "cross_reference_ids": [] }, { "id": "S07-Q001", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "例题1", "problem": "已知：34+86≡7+14≡21≡2+1≡3 (mod 9)，702+20≡2+6≡8≡1 (mod 7)，仿照此形式，利用同余记号书写过程，计算下列各题最终的余数：(1)121+34 (mod 9)；(2)234−117 (mod 11)；(3)432×86 (mod 7)。", "source_answer": "(1)2；(2)7；(3)3。", "source_solution": "(1) 121≡4、34≡7 (mod 9)，4+7=11≡2；(2) 234≡3、117≡7 (mod 11)，3−7=−4≡7；(3) 432≡5、86≡2 (mod 7)，5×2=10≡3。", "topics": [ "同余的加减乘运算", "余数计算" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "无明显重复。", "review": { "answer_status": "verified", "verified_answer": "(1) 2；(2) 7；(3) 3。", "solution_status": "consistent", "note": "原页第1页红笔实际写的是3−7，推得−4≡7；此前题库转录误把手写7识读成1，现已纠正，来源解法本身正确。", "verified_solution": "(1) 模9：121≡4、34≡7，所以4+7=11≡2。(2) 模11：234≡3、117≡7，所以3−7=−4≡7。(3) 模7：432≡5、86≡2，所以5×2=10≡3。" }, "cross_reference_ids": [] }, { "id": "S07-Q002", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "练习1", "problem": "已知：2023≡13 (mod m)，请问m有多少种可能的取值？", "source_answer": "16种。", "source_solution": "m | (2023−13)=2010；2010=2×3×5×67，所以τ(2010)=2^4=16。", "topics": [ "同余定义与整除关系", "约数个数" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [ "题面未说明模数m是否要求大于1。若按正整数模数并允许m=1，则m为2010的16个正因数；若规定m>1，则排除1后为15种。建议明确模数范围。" ], "duplicate_notes": "无明显重复。", "review": { "answer_status": "conditional", "verified_answer": "若允许正模数m=1，共16种；若要求m>1，则15种。", "solution_status": "incomplete", "note": "m需为2010的正因数；来源用τ(2010)=16，隐含把m=1计入。题面未明示是否允许模1，因此保留条件化答案。", "verified_solution": "2023≡13 (mod m)等价于m整除2023−13=2010。2010=2×3×5×67，共有2^4=16个正因数。如果模数允许1，答案16；若教材规定模数必须大于1，则去掉因数1，答案15。" }, "cross_reference_ids": [] }, { "id": "S07-Q003", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "例题2", "problem": "(1)算式1357×2468−321×654被11除，所得余数是多少？被13除所得的余数是多少？(2)多位数A=123412341234⋯1234（共5678个1234），这个数除以25的余数是多少？这个数除以9的余数是多少？除以99的余数是多少？", "source_answer": "(1)除以11余6，除以13余6；(2)除以25余9，除以9余8，除以99余26。", "source_solution": "(1) 模11：1357≡4、2468≡4、321≡2、654≡5，所以余6；模13：1357≡5、2468≡11、321≡9、654≡4，所以5×11−9×4=19≡6。(2) 模25时A的末两位是34，故A≡34≡9；模9时每个1234块的数位和为10，5678个块贡献5678×10≡8；模99时10^4≡1且1234≡46、5678≡35，所以A≡46×35=1610≡26。", "topics": [ "同余的加减乘运算", "多位数与分块同余", "数字和判定法" ], "page_refs": { "problem": [ 1, 2 ], "answer": [ 1, 2 ], "solution": [ 1, 2 ] }, "ambiguities": [], "duplicate_notes": "无明显重复。", "review": { "answer_status": "verified", "verified_answer": "(1) 模11余6、模13余6；(2) 模25余9、模9余8、模99余26。", "solution_status": "incomplete", "note": "各余数复算正确。原页模25手写是A≡34≡9（看末两位）；题库旧转述“每个1234块模25余9”不准确，现已按原页校正并补位置值理由。", "verified_solution": "(1) 模11：1357≡4、2468≡4、321≡2、654≡5，因此4×4−2×5=6。模13：对应余数为5、11、9、4，因此5×11−9×4=19≡6。(2) A是5678个1234依次拼成的数。模25时10^4≡0，所以除最右边的1234外，前面各块都贡献0；末两位34≡9，因此余9。模9时每块数字和10≡1，5678≡8，所以余8。模99时10^4≡1，故A≡5678×1234≡35×46=1610≡26。" }, "cross_reference_ids": [] }, { "id": "S07-Q004", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "练习2", "problem": "A=202620262026⋯2026（共2025个2026），A被7除所得余数是多少？被9除所得余数是多少？", "source_answer": "除以7余0，除以9余0。", "source_solution": "页面写作“202620262026≡0 (mod7)，A≡0 (mod7)”；又写A≡(2+0+2+6)×2025≡0 (mod9)。", "topics": [ "多位数与分块同余", "周期性与等比数列同余", "数字和判定法" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "无明显重复。", "review": { "answer_status": "verified", "verified_answer": "被7除余0；被9除余0。", "solution_status": "incomplete", "note": "原题为四位数块2026重复2025次；按块长4的周期独立计算，模7、模9的余数均为0。", "verified_solution": "令A为2026重复2025次。模7时2026≡3、10^4≡4、10^8≡2，因此连续三块202620262026≡3(1+4+2)=0 (mod 7)。2025=675×3，所有块可分成675组，每组模7都余0。模9时每个2026块的数位和为2+0+2+6=10≡1，2025个块合计余2025≡0。" }, "cross_reference_ids": [] }, { "id": "S07-Q005", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "例题3", "problem": "一个大于1的自然数去除300、243、205时，得到相同的余数，则这个自然数是多少？", "source_answer": "19。", "source_solution": "300≡243≡205 (mod m)，所以m|57且m|38，(57,38)=19，故m=19。", "topics": [ "同余与最大公约数", "相同余数问题" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "与练习3、作业1同属相同余数模型，但数据不同，不是重复题。", "review": { "answer_status": "verified", "verified_answer": "19。", "solution_status": "consistent", "note": "m>1且m同时整除57与38，只能取gcd(57,38)=19；直接验算三数除19余15。", "verified_solution": "相同余数意味着m整除任意两数之差：300−243=57，243−205=38。两差的最大公因数gcd(57,38)=19。因为m>1，唯一可能是m=19；验算300、243、205除以19都余15，满足题意。" }, "cross_reference_ids": [ "XR-06" ] }, { "id": "S07-Q006", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "练习3", "problem": "一个大于1的自然数去除276、317、371时，得余数分别为a、a+7、a+10，求这个自然数。", "source_answer": "17（且a=4）。", "source_solution": "276≡317−7≡371−10 (mod m)，故m|34、m|51，(34,51)=17，m=17；276÷17余4，所以a=4。", "topics": [ "同余与最大公约数", "带参数余数问题" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [ "第2页题面为余数a、a+7、a+10；m=17时371余14，源红笔未核验第三个余数，且其写法只利用差值34、51。" ], "duplicate_notes": "无明显重复。", "review": { "answer_status": "verified", "verified_answer": "模数17，且a=4。", "solution_status": "consistent", "note": "差值约束给出m|34且m|51，m>1故m=17；三个实际余数4、11、14符合a、a+7、a+10。", "verified_solution": "设模数为m。余数依次是a、a+7、a+10，所以317−276=41? 更直接比较：276≡317−7，故m整除317−7−276=34；317−7≡371−10，故m整除371−10−(317−7)=51。于是m整除gcd(34,51)=17。题目给m>1，故m=17。直接计算276=17×16+4、317=17×18+11、371=17×21+14，余数正好为4、4+7、4+10，因此a=4。" }, "cross_reference_ids": [ "XR-06" ] }, { "id": "S07-Q007", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "例题4", "problem": "已知35、1200、103被某自然数除之后所得的余数分别是a−1、a²−3、2a，求这个自然数。", "source_answer": "31。", "source_solution": "35≡a−1即36≡a (mod m)，1200≡a²−3即1203≡a²，103≡2a。于是m|(36²−1203)=93且m|(103−36×2)=31，(93,31)=31，故m=31。", "topics": [ "同余与最大公约数", "参数余数与代数恒等式" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [ "原解推得m=31后未明确求a并回代；补算a=5，三个余数分别为4、22、10，均符合题面。" ], "duplicate_notes": "无明显重复。", "review": { "answer_status": "verified", "verified_answer": "除数31；对应a=5。", "solution_status": "incomplete", "note": "m|93且m|31，故m=1或31；m=1不可能产生题面三个非负余数。m=31时35、1200、103分别余4、22、10，对应a=5，a²−3=22，2a=10。此前复核把1200的余数误写为19，现已纠正。", "verified_solution": "由35≡a−1得a≡36 (mod m)；由1200≡a²−3得a²≡1203 (mod m)，所以m整除36²−1203=93。又103≡2a且a≡36，故m整除103−72=31。因此m整除gcd(93,31)=31。m=1时题面要求的余数a²−3无法为非负余数，排除；只能m=31。此时35除31余4，所以a−1=4、a=5；再验算1200除31余22=a²−3，103除31余10=2a，三项完全符合。" }, "cross_reference_ids": [] }, { "id": "S07-Q008", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "例题5", "problem": "(1) 2025^1000被19除，所得的余数是多少？(2) 在1～2026的所有正整数中，共有多少个整数x，使3^(3x+1)和x^3被5除的余数相同？", "source_answer": "(1)11；(2)405个。", "source_solution": "(1) 2025≡11 (mod 19)，11、11²≡7、11³≡1，周期为3，1000除以3余1，故余数11。(2) 原页化为3^(3x+1)≡3×2^x (mod 5)，列x=1至20的周期表，匹配x=1、12、18、19。原页末尾写2023÷20=101…3；题面范围为1～2026，应按2026÷20=101…6重算，余下6项仍只有x≡1匹配，答案仍为405。", "topics": [ "幂的周期性", "同余方程与周期计数", "模素数运算" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [ "原解析末尾把题面上界2026写作2023进行除法；正确的余数段应为2021—2026，但该段与原解析采用的首3项同样仅有1个匹配，最终答案405不变。" ], "duplicate_notes": "无明显重复。", "review": { "answer_status": "verified", "verified_answer": "(1) 11；(2) 405个。", "solution_status": "error", "note": "题面第二个式子是x³，不是3^x。原答案405正确；源解末尾误用2023÷20=101余3而题面上界是2026。按2026重算为101个整周期加6项，仍只有x≡1匹配，故总数仍405。此前将x³误读成3^x并判0，已撤销。", "verified_solution": "(1) 2025≡11 (mod 19)，且11³≡1，所以2025^1000≡11^(1000 mod3)=11。(2) 模5下3^(3x+1)随x以4为周期变化，x³随x以5为周期变化，比较两列的共同周期20，满足条件的x模20为1、12、18、19，每20个数有4个。2026=101×20+6，前2020个数共有101×4=404个；余下x=2021到2026对应模20的1到6，只有x≡1匹配，再加1个。总数405。", "source_answer": "(1)11；(2)405个。" }, "cross_reference_ids": [] }, { "id": "S07-Q009", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "例题6", "problem": "19^93+93^19的末位数字是几？", "source_answer": "6。", "source_solution": "19^93≡9^93≡9 (mod10)；9^3≡9、9^4≡1，93≡1 (mod4)，故93^19≡3^19≡3^3≡7 (mod10)；和为9+7≡6。", "topics": [ "幂的末位数字", "幂的周期性" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "无明显重复。", "review": { "answer_status": "verified", "verified_answer": "6。", "solution_status": "consistent", "note": "19^93末位为9，93^19末位为7，和末位为6。", "verified_solution": "看个位即可。19的个位是9，9的个位循环为9、1；93次方的个位仍为9。93的个位是3，3的个位循环为3、9、7、1；19除以4余3，所以93^19的个位为7。9+7=16，原数末位为6。" }, "cross_reference_ids": [] }, { "id": "S07-Q010", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "拓展1", "problem": "有一串数：1,1,2,3,5,8,13,21,34,55,89,144,233,……这串数按规律一直写下去，则第2022个数被7除的余数是几？末位数字是几？", "source_answer": "被7除余1，末位数字6。", "source_solution": "按斐波那契数列，模7周期16，2022≡6 (mod16)，a2022≡a6=8≡1 (mod7)；末位数字周期60，2022≡42 (mod60)，a2022=a42≡6 (mod10)。", "topics": [ "斐波那契数列", "周期性", "模7与末位数字" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "无明显重复。", "review": { "answer_status": "verified", "verified_answer": "第2022项被7除余1，末位数字为6。", "solution_status": "consistent", "note": "从F1=F2=1开始核算：模7周期16，模10周期60；2022分别余6、42。", "verified_solution": "令F1=F2=1。斐波那契数列模7的周期为16，2022除以16余6，因此F2022≡F6=8≡1 (mod 7)。末位数字按模10的周期60重复，2022除以60余42；F42=267914296，个位为6。" }, "cross_reference_ids": [] }, { "id": "S07-Q011", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "拓展2", "problem": "2023年元旦是星期日，以后的第47^(37^23)天是星期几？", "source_answer": "星期五。", "source_solution": "47^(37^23)≡5^(37^23) (mod7)；5^6≡1 (mod7)，37^23≡1^23≡1 (mod6)，故天数≡5^1≡5 (mod7)，从星期日顺延5天为星期五。", "topics": [ "幂的周期性", "星期与模7" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "无明显重复。", "review": { "answer_status": "verified", "verified_answer": "星期五。", "solution_status": "consistent", "note": "星期按7天循环；47≡5 (mod7)，且37^23≡1 (mod6)，所以偏移天数模7为5。", "verified_solution": "需要计算N=47^(37^23)除以7的余数。47≡5 (mod7)，而5^6≡1 (mod7)，所以只需算指数模6。37≡1 (mod6)，故37^23≡1 (mod6)，N≡5^1≡5 (mod7)。从星期日往后数5天是星期五。" }, "cross_reference_ids": [] }, { "id": "S07-Q012", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "作业1", "problem": "有一个整数，除39、51、147所得的余数都是3，求这个数。", "source_answer": "m=4、6、12（源解将所求整数理解为除数m）。", "source_solution": "39≡51≡147≡3 (mod m)，所以m分别整除36、12、96，故m|12；又余数3要求m>3，得到m=4、6、12。", "topics": [ "相同余数问题", "最大公约数与约数枚举" ], "page_refs": { "problem": [ 6 ], "answer": [ 6 ], "solution": [ 6 ] }, "ambiguities": [ "第6页题面未明确要求的是除数还是被除数；源解把未知量写作m并列出除数。" ], "duplicate_notes": "与例题3、练习3同为相同余数模型，不是同文档重复题。", "review": { "answer_status": "verified_with_caveat", "verified_answer": "按正整数除数理解：m=4、6、12。", "solution_status": "consistent", "note": "原题措辞省略“这个整数作除数”；结合余数表达式和源解按正除数m理解。三个候选均直接满足。", "verified_solution": "设正整数除数为m。余数为3要求m>3；又m须整除39−3=36、51−3=48和147−3=144，因此m整除gcd(36,48,144)=12。大于3的12的正因数只有4、6、12。逐一代入，39、51、147除以这三个数时余数都为3。" }, "cross_reference_ids": [ "XR-06" ] }, { "id": "S07-Q013", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "作业2", "problem": "有一个整数，用它去除70、110、160所得的3个余数之和是50，那么这个整数是多少？", "source_answer": "29。", "source_solution": "m整除70+110+160−50=290；290=2×5×29，源解列候选29、58并称“验证得m=29”。", "topics": [ "余数和", "约数枚举与整除" ], "page_refs": { "problem": [ 6 ], "answer": [ 6 ], "solution": [ 6 ] }, "ambiguities": [], "duplicate_notes": "无明显重复。", "review": { "answer_status": "verified", "verified_answer": "29。", "solution_status": "incomplete", "note": "m整除290且余数和为50。源解只提到29、58，漏查145、290；逐一计算后只有29满足。", "verified_solution": "设正整数除数为m。三个余数之和为50，所以m整除70+110+160−50=290。每个余数都不超过m−1，故3(m−1)≥50，m≥18。290的正因数中不小于18的只有29、58、145、290。分别求余数和：m=29时12+23+15=50；m=58时12+52+44=108；m=145时70+110+15=195；m=290时70+110+160=340。只有m=29符合。" }, "cross_reference_ids": [] }, { "id": "S07-Q014", "source_id": "S07", "source_filename": "26数论模块训练营第五讲解析(1).pdf", "lesson_or_test": "数论方法训练营第5讲：同余与运算", "source_question_label": "作业3", "problem": "求2^6443÷19的余数。", "source_answer": "10。", "source_solution": "页面列出2^n模19的18项周期：2,4,8,16,13,7,14,9,18,17,15,11,3,6,12,5,10,1；6443÷18余17，故2^6443≡2^17≡10 (mod19)。", "topics": [ "幂的周期性", "模19余数" ], "page_refs": { "problem": [ 6 ], "answer": [ 6 ], "solution": [ 6 ] }, "ambiguities": [], "duplicate_notes": "无明显重复。", "review": { "answer_status": "verified", "verified_answer": "10。", "solution_status": "consistent", "note": "2^6443 mod19独立快速幂为10；6443=18×357+17，且模19的2次幂序列长度为18。", "verified_solution": "模19依次计算2的幂，余数从2开始循环：2、4、8、16、13、7、14、9、18、17、15、11、3、6、12、5、10、1；周期为18。6443=18×357+17，所以2^6443≡2^17≡10 (mod 19)。" }, "cross_reference_ids": [] }, { "id": "S08-Q001", "source_id": "S08", "source_filename": "数论模块班小测3解析版(1).pdf", "lesson_or_test": "数论模块班小测3", "source_question_label": "1", "problem": "（4分）180有多少个因数？有多少个偶数因数？所有因数的和是多少？", "source_answer": "18个，12个，546", "source_solution": "原文未提供推导，仅给出答案：18个，12个，546。", "topics": [ "因数个数", "偶数因数", "约数和", "质因数分解" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [] }, "ambiguities": [], "duplicate_notes": "未发现与本页其他题重复。", "review": { "answer_status": "verified", "verified_answer": "18个因数，12个偶数因数，因数和546", "solution_status": "not_provided", "note": "180=2²×3²×5，因数个数为(2+1)(2+1)(1+1)=18；偶数因数为2×3×2=12；因数和=(1+2+4)(1+3+9)(1+5)=546，均正确。", "verified_solution": "180=2²×3²×5。每个因数可分别选择2、3、5的指数，因此因数个数为(2+1)(2+1)(1+1)=18。偶数因数的2的指数只能为1或2，所以有2×3×2=12个。所有因数的和为(1+2+4)(1+3+9)(1+5)=7×13×6=546。" }, "cross_reference_ids": [] }, { "id": "S08-Q002", "source_id": "S08", "source_filename": "数论模块班小测3解析版(1).pdf", "lesson_or_test": "数论模块班小测3", "source_question_label": "2", "problem": "（4分）一个数除2024余数是4，这个数有多少种情况？", "source_answer": "12-3=9个", "source_solution": "原文未提供推导，仅给出答案：12-3=9个。", "topics": [ "带余除法", "因数枚举", "约数个数" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [] }, "ambiguities": [ "第1页题面‘一个数除2024余数是4’语序不够明确；若不明确‘一个数’指除数，可能产生不同理解。按‘2024除以一个数余4’核验。" ], "duplicate_notes": "未发现与本页其他题重复。", "review": { "answer_status": "verified_with_caveat", "verified_answer": "按“2024除以除数d余4”理解为9个。", "solution_status": "not_provided", "note": "题面中文语序不够明确；按通常题意d>4且d|2020，2020=2²×5×101共有12个正因数，去掉1、2、4得9个。", "verified_solution": "按“2024除以一个正整数d，余数为4”理解，必须有d>4且d整除2024−4=2020。2020=2²×5×101，共有(2+1)(1+1)(1+1)=12个正因数；其中1、2、4不大于4，不能作除数，因此有12−3=9种。原题句式仍有歧义，若不是这个意思需另核。" }, "cross_reference_ids": [] }, { "id": "S08-Q003", "source_id": "S08", "source_filename": "数论模块班小测3解析版(1).pdf", "lesson_or_test": "数论模块班小测3", "source_question_label": "3", "problem": "（4分）一个自然数有12个因数，那么这个数最小是多少？", "source_answer": "60", "source_solution": "原文仅标注【答案】60，未提供推导。", "topics": [ "因数个数", "最小值问题", "质因数分解" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [] }, "ambiguities": [], "duplicate_notes": "未发现与本页其他题重复。", "review": { "answer_status": "verified", "verified_answer": "60", "solution_status": "not_provided", "note": "12的指数分解对应形式中，最小候选为2²×3×5=60；其因数个数为(2+1)(1+1)(1+1)=12。其他形式2¹¹=2048、2⁵×3=96、2³×3²=72均更大，答案正确。", "verified_solution": "若一个数的质因数分解是 p₁^a₁×p₂^a₂×…，因数个数是(a₁+1)(a₂+1)…，要等于12。指数加一的乘法分组只有12、6×2、4×3、3×2×2。各组用最小质数搭配后得到的最小数分别是2¹¹=2048、2⁵×3=96、2³×3²=72、2²×3×5=60。最小的是60。" }, "cross_reference_ids": [] }, { "id": "S08-Q004", "source_id": "S08", "source_filename": "数论模块班小测3解析版(1).pdf", "lesson_or_test": "数论模块班小测3", "source_question_label": "4", "problem": "（4分）分数24/55，27/40的最大公因数和最小公倍数分别是多少？", "source_answer": "", "source_solution": "【解析】3/440，216/5（未给出计算步骤）。", "topics": [ "分数的最大公因数", "分数的最小公倍数", "最大公因数与最小公倍数" ], "page_refs": { "problem": [ 1 ], "answer": [], "solution": [ 1 ] }, "ambiguities": [ "题目未明示所采用的‘分数最大公因数/最小公倍数’定义；核验采用小学竞赛中常用的‘分子取最大公因数、分母取最小公倍数’及其反向定义。" ], "duplicate_notes": "未发现与本页其他题重复。", "review": { "answer_status": "verified_from_solution", "verified_answer": "分数最大公因数=3/440；分数最小公倍数=216/5。", "solution_status": "incomplete", "note": "来源没有独立答案栏，数值由来源解析给出。核验采用小学竞赛常用的分数最大公因数/最小公倍数定义；来源只列数值、未给计算步骤，已补独立推导。题目未明示该定义约定，继续保留说明。", "verified_solution": "按常用竞赛约定，两个分数的最大公因数=分子最大公因数/分母最小公倍数，最小公倍数=分子最小公倍数/分母最大公因数。于是最大公因数为gcd(24,27)/lcm(55,40)=3/440；最小公倍数为lcm(24,27)/gcd(55,40)=216/5。题目未说明采用哪种分数约定，结果依赖此约定。" }, "cross_reference_ids": [] }, { "id": "S08-Q005", "source_id": "S08", "source_filename": "数论模块班小测3解析版(1).pdf", "lesson_or_test": "数论模块班小测3", "source_question_label": "5", "problem": "（4分）求出下列不定方程的正整数解。\n（1）xy+x+y=20\n（2）m²−n²=24", "source_answer": "（1）x=2，6；y=6，2。\n（2）m=7，5；n=5，1。", "source_solution": "原文仅提供答案，未给出推导。", "topics": [ "不定方程", "因式分解", "平方差" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [] }, "ambiguities": [], "duplicate_notes": "同一题含两个独立方程小问，按一组原题计数；未发现与其他题重复。", "review": { "answer_status": "verified", "verified_answer": "（1）(x,y)=(2,6)、(6,2)；（2）(m,n)=(7,5)、(5,1)", "solution_status": "not_provided", "note": "（1）(x+1)(y+1)=21，正整数解仅对应因子对(3,7)、(7,3)，得(2,6)、(6,2)；（2）(m−n)(m+n)=24，取同奇偶正因子对(2,12)、(4,6)，得(m,n)=(7,5)、(5,1)，均正确。", "verified_solution": "（1）xy+x+y=20，两边加1得(x+1)(y+1)=21。因x、y是正整数，x+1、y+1都至少为2；只能取因数对3×7或7×3，所以(x,y)=(2,6)或(6,2)。（2）m²−n²=(m−n)(m+n)=24，且m>n>0，所以两个因数为同奇偶的正整数，较小者在前。可行因数对只有(2,12)、(4,6)，代回得(m,n)=(7,5)、(5,1)。" }, "cross_reference_ids": [] }, { "id": "S09-Q001", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "例题1", "problem": "解下列同余方程，写出正整数通解：（1）4(x−2)≡21−3(5−2x)(mod 11)；（2）(3x+7)/17≡2023(mod 13)；（3）2021x−2626≡2000x+2567(mod 36)。", "source_answer": "（1）x=4+11k，k=0,1,2,…；（2）x=4+13k，k=0,1,2,…；（3）x=9+12k，k=0,1,2,…", "source_solution": "（1）9x≡3(mod11)，45x≡15(mod11)，x≡4(mod11)；（2）3x+7≡8×4(mod13)，3x≡12(mod13)，x≡4(mod13)；（3）21x≡9(mod36)，7x≡3(mod12)，49x≡21(mod12)，x≡9(mod12)。", "topics": [ "一次同余方程", "同余式化简与逆元" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "（1）x=4+11k；（2）x=4+13k；（3）x=9+12k，k=0,1,2,…。", "solution_status": "consistent", "note": "三组通解分别代回原同余式均成立；第（2）式按17在模13下取逆元（17≡4，4^{-1}≡10）核验得x≡4。", "verified_solution": "（1）展开得9x≡3(mod11)，乘5得x≡4(mod11)。（2）17在模13下可逆，乘回17后得3x+7≡2023×17≡6，故3x≡12，x≡4(mod13)。（3）移项得21x≡9(mod36)，两边及模数同除以3，得7x≡3(mod12)；7的逆元为7，故x≡9(mod12)。列出正整数通解即可。" }, "cross_reference_ids": [] }, { "id": "S09-Q002", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "练习1", "problem": "解下列同余方程，写出正整数x的通解及最小值：4(x−2)≡21−3(5−2x)(mod 17)；7x≡2x+9(mod 13)。", "source_answer": "第一题x=10+17k，k=0,1,2,…；第二题x=7+13k，k=0,1,2,…；最小值分别为10、7。", "source_solution": "第一题原页：15x≡14(mod17)，60x≡56(mod17)，9x≡5(mod17)，18x≡10(mod17)，x≡10(mod17)；第二题：5x≡9(mod13)，40x≡72(mod13)，x≡7(mod13)。", "topics": [ "一次同余方程", "正整数通解与最小值" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "第一小题与例题1（1）结构相同但模数不同，不视为重复题。", "review": { "answer_status": "verified", "verified_answer": "第一题x=10+17k，最小正整数10；第二题x=7+13k，最小正整数7；k=0,1,2,…。", "solution_status": "consistent", "note": "两解与最小正整数均正确；放大原页确认首步为15x≡14，修正题库旧转录的18x笔误。", "verified_solution": "第一题化简为−2x≡14(mod17)，等价于15x≡14；乘4得60x≡56，再化为9x≡5。乘2得18x≡10，所以x≡10(mod17)。原页首步为15x≡14，题库旧转录误作18x≡14，现已校正。第二题化为5x≡9(mod13)，乘8得x≡72≡7(mod13)。" }, "cross_reference_ids": [] }, { "id": "S09-Q003", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "例题2", "problem": "用同余法解不定方程（写出正整数解）：29x+11y=300；23x−13y=1000。", "source_answer": "第一式x=2，y=22；第二式x=48+13k，y=8+23k，k=0,1,2,…。", "source_solution": "第一式：7x≡3(mod11)，56x≡24(mod11)，x≡2(mod11)，得{x=2,y=22}。第二式：10x≡1000(mod13)，x≡100(mod13)，x≡9(mod13)，得{x=48+13k，y=8+23k}，k=0,1,2,…。", "topics": [ "一次不定方程", "同余法求正整数解" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "（1）唯一正整数解(x,y)=(2,22)；（2）x=48+13k、y=8+23k，k=0,1,2,…。", "solution_status": "consistent", "note": "第一式的正整数解确实只有(2,22)；第二式代回参数式恒等成立，k≥0时均为正整数。", "verified_solution": "（1）模11化为7x≡3，故x≡2(mod11)。令x=2+11t，则y=22−29t；要求x、y为正整数，只能t=0，得到(2,22)。（2）模13有10x≡1000≡12，故x≡9(mod13)。写x=9+13t，代回得y=−61+23t；正整数条件要求t≥3。令k=t−3，即得x=48+13k、y=8+23k。" }, "cross_reference_ids": [] }, { "id": "S09-Q004", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "练习2", "problem": "用同余法解不定方程（写出正整数解）：71x+17y=800；152x−56y=2024。", "source_answer": "第一式x=6，y=22；第二式x=17+7k，y=10+19k，k=0,1,2,…。", "source_solution": "第一式：3x≡1(mod17)，18x≡6(mod17)，x≡6(mod17)，得{x=6,y=22}。第二式：19x−7y=253，5x≡1(mod7)，15x≡3(mod7)，x≡3(mod7)，得{x=17+7k，y=10+19k}，k=0,1,2,…。", "topics": [ "一次不定方程", "同余法求正整数解" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "（1）唯一正整数解(x,y)=(6,22)；（2）x=17+7k、y=10+19k，k=0,1,2,…。", "solution_status": "consistent", "note": "两组参数解代回均成立，且覆盖全部正整数解。", "verified_solution": "（1）模17得3x≡1，所以x≡6(mod17)。令x=6+17t，代回得y=22−71t；正整数条件只允许t=0。（2）原式除以8得19x−7y=253。模7得5x≡1，所以x≡3(mod7)。写x=3+7t，代回得y=−28+19t；正整数条件要求t≥2。令k=t−2，得到x=17+7k、y=10+19k。" }, "cross_reference_ids": [] }, { "id": "S09-Q005", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "例题3", "problem": "解下列同余方程组：（1）x≡2(mod3)，x≡3(mod11)；（2）x≡7(mod3)，x≡3(mod10)，x≡6(mod13)。", "source_answer": "（1）x=14+33k，k=0,1,2,…；（2）x=253+390k，k=0,1,2,…。", "source_solution": "（1）试算表列出x模3余2、模11余3，取x=14，故x=14+33k。 （2）逐项试算6,19,58,97,136,175,214,253，取x=253，故x=253+390k。", "topics": [ "同余方程组", "中国剩余定理（互质模数）" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "（1）x=14+33k；（2）x=253+390k；两题均取k=0,1,2,…。", "solution_status": "consistent", "note": "14和253分别满足全部同余条件，模数乘积分别为33和390。", "verified_solution": "（1）x≡2(mod3)，且x≡3(mod11)。在3、14、25……中，14是第一个也就是模33的解，故x=14+33k。（2）x≡7(mod3)等价于x≡1(mod3)。x≡3(mod10)、x≡6(mod13)并合并后，最小正解为253；因为3、10、13两两互质，所有解相差390，故x=253+390k。" }, "cross_reference_ids": [] }, { "id": "S09-Q006", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "练习3", "problem": "解同余方程组：6x+3≡0(mod9)，3x−4≡0(mod5)，4x+1≡0(mod7)。", "source_answer": "x=105k−2，k=1,2,3,…。", "source_solution": "化为x≡1(mod3)，x≡3(mod5)，x≡5(mod7)，得x+2=[3,5,7]k，故x=105k−2，k=1,2,3,…。", "topics": [ "同余方程组", "中国剩余定理（互质模数）" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "x=105k−2，k=1,2,3,…（等价于x=103+105t，t=0,1,2,…）。", "solution_status": "consistent", "note": "105k−2对模3、5、7分别余1、3、5，且最小正解为103。", "verified_solution": "第一式6x+3≡0(mod9)化为x≡1(mod3)；第二式3x−4≡0(mod5)化为x≡3(mod5)；第三式4x+1≡0(mod7)化为x≡5(mod7)。逐一检验得最小正解103；模数3、5、7两两互质，解每隔105重复，所以x=103+105t，t≥0，也就是x=105k−2，k≥1。" }, "cross_reference_ids": [] }, { "id": "S09-Q007", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "例题4", "problem": "一个数除以3、5、7、11的余数分别是2、3、4、5，求符合条件的最小的数。", "source_answer": "x_min=368。", "source_solution": "2x−1=[3,5,7]=105，x=53；x=53+105k≡5(mod11)，6k≡1(mod11)，k≡3(mod11)，所以x=368+1155m，m=0,1,2,…，x_min=368。", "topics": [ "同余方程组", "最小正整数与中国剩余定理" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "与例题5（1）同为余数递增型，但模数和余数不同，不视为重复。", "review": { "answer_status": "verified", "verified_answer": "最小符合条件的数为368。", "solution_status": "consistent", "note": "368除以3、5、7、11的余数为2、3、4、5。", "verified_solution": "前三个条件可合并：2n−1同时被3、5、7整除，故2n−1=105k，n=53+105t。再要求n≡5(mod11)：53+105t≡9+6t≡5，所以6t≡7(mod11)，t≡3(mod11)。最小t=3，得n=368；之后每隔1155重复。" }, "cross_reference_ids": [] }, { "id": "S09-Q008", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "练习4", "problem": "在200至300之间，有三个连续的自然数，其中，最小的能被3整除，中间的能被7整除，最大的能被13整除，那么这样的三个连续自然数分别是多少？", "source_answer": "258，259，260。", "source_solution": "设n,n+1,n+2；筛选n=11,24,63,102,141,180,219,258，满足n≡0(mod3)、n≡6(mod7)、n≡11(mod13)，故为258,259,260。", "topics": [ "同余方程组", "连续自然数整除问题" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "258、259、260。", "solution_status": "consistent", "note": "258、259、260分别满足3、7、13的整除条件，且均在200至300之间。", "verified_solution": "令三数为n、n+1、n+2。条件变为n≡0(mod3)、n≡6(mod7)、n≡11(mod13)。在200至298之间逐步筛选，唯一满足三式的是n=258，因此三数为258、259、260。" }, "cross_reference_ids": [] }, { "id": "S09-Q009", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "例题5", "problem": "（1）一个数除以7、9、11、13分别余6、7、8、9，求符合条件的最小数；（2）有三个连续自然数，其中最小的能被15整除，中间的能被17整除，最大的能被19整除，请写出一组这样的三个连续自然数，并使得总和最小。", "source_answer": "（1）n_min=4507；（2）2430、2431、2432。", "source_solution": "（1）2n−5=[7,9,11,13]=9009，n_min=(9009+5)÷2=4507。（2）设n,n+1,n+2，化为2n≡15(mod15)、2n≡15(mod17)、2n≡15(mod19)，故2n−15=[15,17,19]=4845，n_min=2430，三数为2430、2431、2432。", "topics": [ "同余方程组", "最小值问题", "连续自然数整除问题" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "（1）与例题4同属余数型同余方程组，但题面不同；（2）与练习4同属连续数整除问题，但题面不同。", "review": { "answer_status": "verified", "verified_answer": "（1）最小数4507；（2）最小和的一组三数为2430、2431、2432。", "solution_status": "consistent", "note": "（1）4507对7、9、11、13余6、7、8、9；（2）2430、2431、2432分别被15、17、19整除条件满足。", "verified_solution": "（1）设数为n。四个条件分别给出2n−5被7、9、11、13整除，所以2n−5是它们最小公倍数9009的倍数。要使n为整数，倍数因子须为奇数；最小取1，n=(9009+5)/2=4507。（2）令三数为n、n+1、n+2，则n≡0(mod15)、n≡16(mod17)、n≡17(mod19)。因此2n−15同时被15、17、19整除，最小公倍数为4845。取最小可使n为整数的正奇数倍1，得n=(4845+15)/2=2430；三数和随n增大，故这是总和最小的一组。" }, "cross_reference_ids": [] }, { "id": "S09-Q010", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "例题6", "problem": "数2026^2027除以231的余数是多少？", "source_answer": "40。", "source_solution": "231=3×7×11；2026^2027≡1(mod3)，≡3^2027≡3^5≡5(mod7)，≡2^2027≡2^7≡7(mod11)，由①②③得2026^2027≡40(mod231)，为40。", "topics": [ "幂的同余", "中国剩余定理" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "40。", "solution_status": "consistent", "note": "40模3、7、11分别为1、5、7，故由中国剩余定理余数为40。", "verified_solution": "231=3×7×11。2026^2027模3余1；模7时2026≡3且2027≡5(mod6)，故余数为3^5≡5；模11时2026≡2且2027≡7(mod10)，故余数为2^7≡7。40在模3、7、11下分别余1、5、7，且0≤40<231，所以所求余数为40。" }, "cross_reference_ids": [] }, { "id": "S09-Q011", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "拓展1", "problem": "数2027^2026除以105的余数是多少？", "source_answer": "4。", "source_solution": "105=3×5×7；2027^2026≡2^2026≡2^2≡1(mod3)，≡2^2026≡2^2≡4(mod5)，≡4^2026≡4^1≡4(mod7)，所以2027^2026≡4(mod105)。", "topics": [ "幂的同余", "中国剩余定理" ], "page_refs": { "problem": [ 5 ], "answer": [ 5 ], "solution": [ 5 ] }, "ambiguities": [], "duplicate_notes": "与例题6同属幂的同余，但底数、指数、模数不同，不视为重复。", "review": { "answer_status": "verified", "verified_answer": "4。", "solution_status": "consistent", "note": "4对3、5、7分别余1、4、4，满足三项余数条件。", "verified_solution": "105=3×5×7。2027模3、5、7分别为2、2、4。把这些数分别取2026次幂，按周期得到余数1、4、4。数4同时满足x≡1(mod3)、x≡4(mod5)、x≡4(mod7)，并且在0至104之间，所以所求余数为4。" }, "cross_reference_ids": [] }, { "id": "S09-Q012", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "拓展2", "problem": "数7^9999的最后三位是多少？", "source_answer": "143。", "source_solution": "7^9999≡(−1)^9999≡7(mod8)；7^9999≡(7^10)^999×7^9≡(−1)×107≡18(mod125)；由①②得7^9999≡18+125=143(mod1000)。", "topics": [ "幂的同余", "末三位问题", "分解模数" ], "page_refs": { "problem": [ 5 ], "answer": [ 5 ], "solution": [ 5 ] }, "ambiguities": [], "duplicate_notes": "无重复题。", "review": { "answer_status": "verified", "verified_answer": "143。", "solution_status": "consistent", "note": "独立计算7^9999 mod1000=143；页面列出的模8、模125结果均一致。", "verified_solution": "先算模8：7≡−1，所以7^9999≡−1≡7(mod8)。再算模125：7^10≡−1且7^9≡107，因此7^9999=(7^10)^999×7^9≡−107≡18(mod125)。设末三位数为18+125k；它还须模8余7，即2+5k≡7(mod8)，所以k≡1(mod8)。0至999内对应k=1，得到143。" }, "cross_reference_ids": [] }, { "id": "S09-Q013", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "作业1", "problem": "解同余方程，写出正整数x的通解和最小值：（1）3(x−2)+2x≡4−2(x−1)(mod10)；（2）4(x−1)+7≡6(2x+3)(mod9)。", "source_answer": "（1）x=6+10k，k=0,1,2,…，x_min=6；（2）x=6+9k，k=0,1,2,…，x_min=6。", "source_solution": "（1）7x≡2(mod10)，21x≡6(mod10)，x≡6(mod10)；（2）8x≡3(mod9)，64x≡24(mod9)，x≡6(mod9)。", "topics": [ "一次同余方程", "正整数通解与最小值" ], "page_refs": { "problem": [ 7 ], "answer": [ 7 ], "solution": [ 7 ] }, "ambiguities": [], "duplicate_notes": "与练习1同属一次同余方程，但题面不同。", "review": { "answer_status": "verified", "verified_answer": "（1）x=6+10k，最小正整数6；（2）x=6+9k，最小正整数6；k=0,1,2,…。", "solution_status": "consistent", "note": "独立展开后分别化为7x≡2(mod10)与8x≡3(mod9)，两解类均为x≡6；最小正整数都是6。", "verified_solution": "（1）左边化为5x−6，右边化为6−2x，所以7x≡12≡2(mod10)。7的逆元模10为3，故x≡6(mod10)。（2）展开移项得−8x≡15≡6(mod9)。−8≡1(mod9)，故x≡6(mod9)。由通解可见两题最小正整数均为6。" }, "cross_reference_ids": [] }, { "id": "S09-Q014", "source_id": "S09", "source_filename": "26数论模块训练营第六讲解析(1).pdf", "lesson_or_test": "数论方法训练营第6讲 一次同余方程（组）解法", "source_question_label": "作业2", "problem": "韩信点兵故事：士兵3人一排多出2名，5人一排多出3名，7人一排多出2名；韩信有1500名将士，战斗死伤四五百人，敌人不足五百。根据故事中的条件，算出韩信有多少将士。", "source_answer": "页面手写结论为n=23+105k，且1000a且都是数字，d−a只能为1至8，故d−a=8，再得c−b=9。于是a=1、d=9、b=0、c=9，原数1099。反验9901−1099=8802。" }, "cross_reference_ids": [] }, { "id": "S12-Q004", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "练习2", "problem": "已知 ̅abcd+̅abc+̅ab+a=1370，求̅abcd。", "source_answer": "1234", "source_solution": "1111a+111b+11c+d=1370；先取a=1，再得111b+11c+d=259，取b=2，得11c+d=37，故c=3、d=4。", "topics": [ "位值原理", "数字方程" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "1234。", "solution_status": "consistent", "note": "独立枚举四位数也只有1234；位值展开后逐位限定a、b、c、d，答案唯一。", "verified_solution": "令abcd表示四位数，则abcd+abc+ab+a=1111a+111b+11c+d=1370。若a≥2，首项1111a已超过1370，因此a=1。此时111b+11c+d=259；b只能为2，因为b≤1时左边最多219，b≥3时已超过259。于是11c+d=37，得c=3、d=4，答案1234。" }, "cross_reference_ids": [] }, { "id": "S12-Q005", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "例题3（1）", "problem": "n=̅labcde，m=̅abcdel，且m=3n，求a+b+c+d+e。", "source_answer": "26或25（第2页红笔旁注）", "source_solution": "̅abcdel=3̅labcde；10̅abcde+l=3(100000l+̅abcde)，得̅abcde=42857l。页上红笔列出l=1、̅abcde=42857，l=2、̅abcde=28571。", "topics": [ "位值原理", "循环移位与倍数关系" ], "page_refs": { "problem": [ 2 ], "answer": [], "solution": [ 2 ] }, "ambiguities": [ "来源红笔第二行确写l=2、abcde=28571，但按方程应为85714；旁注“26或25”对应两种正确解，题目本身有两个答案。" ], "duplicate_notes": "", "review": { "answer_status": "non_unique", "verified_answer": "数字和有两种可能：26或25。", "solution_status": "error", "note": "由原式及五位数范围复核，l=1时t=42857，数字和26；l=2时t=85714，数字和25。原页红笔把第二个t误写为28571，但旁注“26或25”正确反映两个解；题目非唯一。", "verified_solution": "令t=overline{abcde}。由overline{abcdel}=3·overline{labcde}得10t+l=3(100000l+t)，所以7t=299999l，t=42857l。t必须是五位数且l为非零首位数字，因此l只能为1或2：l=1时t=42857，数字和26；l=2时t=85714，数字和25。原页红笔第二行误把85714写成28571。题目有两解，不能给唯一数值。" }, "cross_reference_ids": [] }, { "id": "S12-Q006", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "例题3（2）", "problem": "一个四位数̅abcd，满足̅abcd=̅ab×̅cd×2，求̅abcd。", "source_answer": "", "source_solution": "令̅ab=x、̅cd=y，则100x+y=2xy，(2x−1)(y−50)=50；页上取2x−1=25、y−50=2，得x=13、y=52。", "topics": [ "位值原理", "两位数分块", "整数因式分解" ], "page_refs": { "problem": [ 2 ], "answer": [], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "1352。", "solution_status": "consistent", "note": "逐一考察50的奇因数并结合x、y都为两位数，只得x=13、y=52，四位数1352。", "verified_solution": "令x=ab、y=cd，则四位数abcd=100x+y，题设给100x+y=2xy，整理为(2x−1)(y−50)=50。x是两位数，所以2x−1是大于19的奇因数；50的奇因数只有1、5、25，只有25符合，故x=13、y−50=2，即y=52。于是abcd=1352，代回有13×52×2=1352。" }, "cross_reference_ids": [] }, { "id": "S12-Q007", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "练习3", "problem": "一个六位数，它的个位上的数字是7。如果把数字7移到第一位，所得的数是原数的5倍。这个六位数是____。", "source_answer": "142857", "source_solution": "令t=overline{abcde}，原数为10t+7，移位后为700000+t。由700000+t=5(10t+7)得49t=699965，t=14285，故原数为142857。", "topics": [ "位值原理", "循环移位与倍数关系" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [ "第2页红笔推导中间的带横线数字局部潦草，按题面与最终答案保留可辨内容。" ], "duplicate_notes": "与例题3（1）同现142857循环移位结构，但不是同文档重复题。", "review": { "answer_status": "verified", "verified_answer": "142857。", "solution_status": "consistent", "note": "由题意列式得t=14285，原数142857；复乘5得714285，正好把末位7移到首位。", "verified_solution": "设前五位组成t。原数是10t+7，移位后是700000+t。方程700000+t=5(10t+7)化为49t=699965，所以t=14285，原数为142857。检查142857×5=714285，符合“把末位7移到第一位”。" }, "cross_reference_ids": [] }, { "id": "S12-Q008", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "例题4", "problem": "一个四位数\\overline{7ab7}，将它连续重复写2026次，成为一个多位数后能被91整除，那么这个四位数\\overline{7ab7}是____。", "source_answer": "7917或7007。", "source_solution": "设B=\\overline{7ab7}。每三个重复块组成的数为B(10^8+10^4+1)=B×100010001，而100010001=91×1099011。2026=3×675+1，故重复2026次的数模91与B同余；要被91整除，必须91|B。四位数B在7000至7999且末位为7，写作91k时77≤k≤87，且B末位等于k末位，因此k只能为77或87，得到B=7007或7917。", "topics": [ "位值原理", "重复数", "整除性" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [ "扫描页中块的末位7受红笔修改覆盖，文本层曾漏录；放大后的红笔重复式清楚写作7ab7，且答案7917、7007均与四位数条件相符。题库题面按该证据修正。" ], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "7007或7917。", "solution_status": "consistent", "note": "放大原页显示重复块为四位数7ab7；按三块分组，2026=3×675+1，故大数模91与块本身同余。满足91整除、范围7000—7999且末位7的仅7007、7917。旧题库遗漏末位7并误判为3位数，已纠正。", "verified_solution": "令B=7ab7。每三个连续块的值是B×(10^8+10^4+1)，括号等于100010001=91×1099011，因此每组三块都能被91整除。2026个块可分成675组三块再加1块；多位数模91的余数就是B的余数，所以B必须是91的倍数。7000至7999之间的91倍数写成91k，k只能从77到87；又因91k末位与k同余于10，末位为7要求k=77或87，故B=7007或7917。" }, "cross_reference_ids": [] }, { "id": "S12-Q009", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "例题5-①", "problem": "(101)₂×(1011)₂−(11011)₂=____。", "source_answer": "28=(11100)₂", "source_solution": "页面仅给红笔答案28=(11100)₂。", "topics": [ "二进制", "进制运算" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "例题5为同一编号下的五个独立计算小题，按独立题目记录。", "review": { "answer_status": "verified", "verified_answer": "28=(11100)₂。", "solution_status": "consistent", "note": "(101)₂=5、(1011)₂=11、(11011)₂=27，故5×11−27=28=(11100)₂。", "verified_solution": "把二进制数换成十进制：101₂=5，1011₂=11，11011₂=27。计算5×11−27=28，再把28换回二进制得11100₂。" }, "cross_reference_ids": [] }, { "id": "S12-Q010", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "例题5-②", "problem": "(11000111)₂−(10101)₂÷(11)₂=(____)₂。", "source_answer": "(11000000)₂", "source_solution": "页面仅给红笔答案(11000000)₂。", "topics": [ "二进制", "进制运算" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "(11000000)₂。", "solution_status": "consistent", "note": "11000111₂=199，10101₂=21，11₂=3；按先除后减，199−21÷3=192=11000000₂。", "verified_solution": "按运算顺序先算除法：11000111₂=199，10101₂=21，11₂=3，因此原式为199−21÷3=192。192=128+64，写成二进制是11000000₂。" }, "cross_reference_ids": [] }, { "id": "S12-Q011", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "例题5-③", "problem": "(3021)₄+(605)₇=(____)₁₀。", "source_answer": "500", "source_solution": "页面仅给红笔答案500。", "topics": [ "多进制转换", "进制运算" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "500。", "solution_status": "consistent", "note": "3021₄=201，605₇=299，和为500。", "verified_solution": "(3021)₄=3×4³+0×4²+2×4+1=201；(605)₇=6×7²+0×7+5=299。相加为500。" }, "cross_reference_ids": [] }, { "id": "S12-Q012", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "例题5-④", "problem": "(63121)₈−(1247)₈−(16034)₈−(26531)₈−(1744)₈=____。", "source_answer": "(13121)₈。", "source_solution": "页面红笔答案为(13121)₈。独立换算得十进制5713，等于(13121)₈。", "topics": [ "八进制", "进制运算" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "(13121)₈。", "solution_status": "consistent", "note": "放大原页可辨红笔答案为(13121)₈；逐项转换后十进制差为5713，换回八进制为13121₈。旧题库文本转录漏掉了数字1。", "verified_solution": "逐项转成十进制：(63121)₈=26193，(1247)₈=679，(16034)₈=7196，(26531)₈=11609，(1744)₈=996。依次相减得5713。把5713转回八进制：5713=1×8⁴+3×8³+1×8²+2×8+1，因此答案是(13121)₈。" }, "cross_reference_ids": [] }, { "id": "S12-Q013", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "例题5-⑤", "problem": "若(1030)ₙ=140，则n=____。", "source_answer": "5", "source_solution": "1×n³+3n=140，n(n²+3)=140，n=5。", "topics": [ "多进制表示", "进制方程" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "n=5。", "solution_status": "consistent", "note": "n=5时125+15=140；在n≥4范围内左式严格递增，故唯一。", "verified_solution": "按位值展开，(1030)ₙ=n³+3n=140。因为数字3出现在该数中，底数n≥4。检验n=5得125+15=140；函数n³+3n随正整数n增大而严格增大，因此解唯一。" }, "cross_reference_ids": [] }, { "id": "S12-Q014", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "练习5-①", "problem": "567=(____)₈=(____)₅=(____)₂。", "source_answer": "(1067)₈=(4232)₅=(1000110111)₂", "source_solution": "页面红笔依次填写1067、4232、1000110111。", "topics": [ "多进制转换", "八进制", "五进制", "二进制" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "(1067)₈=(4232)₅=(1000110111)₂。", "solution_status": "consistent", "note": "三种表示各自换回十进制都等于567。", "verified_solution": "先将567分别换算。八进制：567=1×8³+0×8²+6×8+7，故为1067₈。五进制：567=4×125+2×25+3×5+2，故为4232₅。二进制：567=512+32+16+4+2+1，故为1000110111₂。" }, "cross_reference_ids": [] }, { "id": "S12-Q015", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "练习5-②", "problem": "在八进制中，1234−456−322=____。", "source_answer": "(234)₈", "source_solution": "页面仅给红笔答案(234)₈。", "topics": [ "八进制", "进制运算" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "(234)₈。", "solution_status": "consistent", "note": "逐项换算并相减得十进制156，换回八进制为234。", "verified_solution": "在八进制中，1234₈=668，456₈=302，322₈=210（以上均为十进制值）。668−302−210=156；156=2×64+3×8+4，所以答案234₈。" }, "cross_reference_ids": [] }, { "id": "S12-Q016", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "练习5-③", "problem": "在九进制中，14438+3123−7120−11770+5766=____。", "source_answer": "(4473)₉", "source_solution": "页面仅给红笔答案(4473)₉。", "topics": [ "九进制", "进制运算" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "partially_incorrect", "verified_answer": "(4438)₉。", "solution_status": "error", "note": "原卷只给答案4473₉，但该值为3306₁₀；算式为3275₁₀=4438₉，确认来源答案错误。", "verified_solution": "逐项转为十进制：14438₉=9836，3123₉=2289，7120₉=5202，11770₉=7920，5766₉=4272。计算9836+2289−5202−7920+4272=3275。再转回九进制：3275=4×9³+4×9²+3×9+8，故为4438₉。原卷红笔答案4473₉换算为3306，和算式不符。" }, "cross_reference_ids": [] }, { "id": "S12-Q017", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "例题6", "problem": "一个八进制数(̅abcb)₈在八进制下满足(̅bcba)₈−(̅abcb)₈=(3164)₈，其中相同字母代表相同数字，不同字母代表不同数字。那么这个八进制数(̅abcb)₈化成十进制数是____。", "source_answer": "1991", "source_solution": "页面列出8+a−b=4、b−c=1，得a=3、b=7、c=0；(3707)₈=3×8³+7×8²+7=1991。", "topics": [ "八进制", "位值原理", "字母代数与进制" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "1991。", "solution_status": "consistent", "note": "遍历八进制数字0—7并代回差式，唯一解(a,b,c)=(3,7,0)，答案1991。", "verified_solution": "设A=(abcb)₈、B=(bcba)₈。由题面枚举八进制数字并满足首位非零且a、b、c互异，只有(a,b,c)=(3,7,0)使B−A=3164₈。于是A=(3707)₈=3×512+7×64+7=1991。回代：(7073)₈−(3707)₈=(3164)₈。" }, "cross_reference_ids": [] }, { "id": "S12-Q018", "source_id": "S12", "source_filename": "26数论模块训练营第八讲解析(1).pdf", "lesson_or_test": "数论方法训练营 第8讲 位值与进制", "source_question_label": "拓展1", "problem": "将4个不同的数字排在一起，可以组成24个不同的四位数。将这24个四位数按从小到大的顺序排列，第二个是5的倍数；按从大到小排列，第二个是不能被4整除的偶数；按从小到大排列的第五个与第二十个的差在3000～4000之间。求这24个四位数中最大的那个数。", "source_answer": "7543", "source_solution": "红笔设1≤a0，设N=2^a×3^b×5^c×……。乘2后为平方、乘3后为立方、乘5后为五次方，分别要求各质因数指数满足相应模条件。对2，a为奇数且被3、5整除，最小a=15；对3，b为偶数、b≡2(mod3)且被5整除，最小b=20；对5，c为偶数、被3整除且c≡4(mod5)，最小c=24；其他质因数指数须同时被2、3、5整除，最小取0。因此正自然数分支为2^15×3^20×5^24。若0属于自然数，N=0也满足三个乘积分别为平方、立方和五次方，且是更小解。" }, "cross_reference_ids": [] }, { "id": "S13-Q008", "source_id": "S13", "source_filename": "26数论模块训练营第九讲解析(1).pdf", "lesson_or_test": "第9讲 三角形数与完全平方数", "source_question_label": "练习4", "problem": "有5个连续自然数，它们的和为一个平方数，中间三个数的和为立方数，则这五个数中最小数的最小值为____。", "source_answer": "1123", "source_solution": "设五数为n−2、n−1、n、n+1、n+2。5n为平方数，故5|n；3n为立方数，故9|n。由两条幂条件合并指数奇偶/模3要求，3²×5³|n，最小n=3²×5³=1125；并验证5n=5625=75²、3n=3375=15³，所以最小数为1123。", "topics": [ "连续自然数", "完全平方数", "完全立方数", "质因数分解" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "1123。", "solution_status": "consistent", "note": "原页放大后确认为3²×5³=1125；旧题库把3²误录为3³，已纠正。", "verified_solution": "设五数为n−2、n−1、n、n+1、n+2，则总和5n是平方数，中间三数和3n是立方数。由5n为平方可知n中5的指数为奇数；由3n为立方可知n中3的指数至少为2。合并两种幂的指数条件，最小n=3²×5³=1125；其他质因数若出现，指数需同时满足平方与立方条件，至少为6，不会使n更小。验证5n=5625=75²，3n=3375=15³。因此最小数n−2=1123。" }, "cross_reference_ids": [] }, { "id": "S13-Q009", "source_id": "S13", "source_filename": "26数论模块训练营第九讲解析(1).pdf", "lesson_or_test": "第9讲 三角形数与完全平方数", "source_question_label": "例题5", "problem": "A是一个两位数，它的6倍是一个三位数B，如果把B放在A的左边或者右边得到两个不同的五位数，并且这两个五位数的差是一个完全平方数（整数的平方），求A的所有可能取值之和。", "source_answer": "145", "source_solution": "B=6A；100B+A=601A，1000A+B=1006A；两数之差405A=9²×5×A，所以A=5×a²。两位数A为5×2²、5×3²、5×4²，即20、45、80；和为20+45+80=145。", "topics": [ "完全平方数", "位值制与数的拼接", "质因数分解" ], "page_refs": { "problem": [ 3 ], "answer": [ 3 ], "solution": [ 3 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "145。", "solution_status": "consistent", "note": "穷举两位A或由405A为平方得A=5k²，候选20、45、80，和145。", "verified_solution": "令两位数为A，B=6A。把B接在A左边得到100B+A=601A；接在A右边得到1000A+B=1006A；两数差为405A=81×5×A。此差为平方数当且仅当5A为平方数，即A=5k²。两位数解为20、45、80，和为145；逐个代回均得到平方差。" }, "cross_reference_ids": [] }, { "id": "S13-Q010", "source_id": "S13", "source_filename": "26数论模块训练营第九讲解析(1).pdf", "lesson_or_test": "第9讲 三角形数与完全平方数", "source_question_label": "练习5", "problem": "有两个两位数，它们的差是14，将它们分别平方，得到的两个平方数的末两位数（个位数和十位数）相同，那么这两个两位数是多少？（请写出所有可能的答案）", "source_answer": "(18,32)、(43,57)、(68,82)", "source_solution": "设两数为n、n+14，则(n+14)²−n²=28(n+7)=100k；化为7(n+7)=25k，故n+7=25、50、75，n=18、43、68，得到(18,32)、(43,57)、(68,82)，均符合。", "topics": [ "平方数", "同余与末位数", "因式分解" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "(18,32)、(43,57)、(68,82)。", "solution_status": "consistent", "note": "独立遍历全部两位数对，只有三组；平方末两位已逐一代回。", "verified_solution": "设较小数为n，较大数为n+14，其中10≤n≤85。两平方末两位相同，等价于其差28(n+7)为100的倍数；化简得7(n+7)=25k。因7与25互质，n+7必须是25的倍数，范围内只有25、50、75，故n=18、43、68。对应平方末两位分别为24、49、24，全部符合。" }, "cross_reference_ids": [] }, { "id": "S13-Q011", "source_id": "S13", "source_filename": "26数论模块训练营第九讲解析(1).pdf", "lesson_or_test": "第9讲 三角形数与完全平方数", "source_question_label": "例题6", "problem": "末三位相同，且不为0的完全平方数有几个？", "source_answer": "3（红笔列出289444、213444、925444）", "source_solution": "红笔解析先取末三位为444，因38²=1444；设m²=1000a+444，则(m−38)(m+38)=1000(a−1)=1000k=125×8k，列出m=538、462、962，得到289444、213444、925444，共3个。", "topics": [ "完全平方数", "同余与末位数" ], "page_refs": { "problem": [ 4 ], "answer": [ 4 ], "solution": [ 4 ] }, "ambiguities": [ "第4页题面未给数值范围；‘末三位相同’也未明确是否要求至少四位数。若隐含限制为某个有限范围，范围未印出。" ], "duplicate_notes": "", "review": { "answer_status": "conditional", "verified_answer": "若补入“六位完全平方数”的范围，答案为 3 个：213444、289444、925444；按现有无限范围题面则有无穷多个。", "solution_status": "incomplete", "note": "原页第4页在题干上方可辨认红笔“六位”。三者答案在六位约束下正确；若只按印刷题面，不设位数上界，则n=38+500k可产生无穷多个平方数，均以444结尾。将手写补充与印刷条件分开保存。 本轮独立复核（原PDF第4页，2026-10-08）：红笔“六位”可辨，印刷题干本身无位数上限；故3个答案仅在接受手写范围时成立。来源还直接写末三位为444，未证明为何111—999中其余重复数字不能为平方末尾；本轮补全模8与模25筛选。", "verified_solution": "设平方数为n²。末三位相同且非0时，末三位只能是111、222、…、999。平方数除以8余0、1或4，先只剩444、777、888；但777除以25余2、888除以25余13，而平方数除以25可能的余数只有0、1、4、6、9、11、14、16、19、21、24，所以末三位只能是444。若接受原页红笔补充“六位”，则316≤n≤999，且n²≡444（mod 1000）。因38²≡444（mod 125），有(n−38)(n+38)≡0（mod 125）；两因数相差76，不可能同时含因数5，所以n≡38或87（mod 125）。又444≡4（mod 8），故n≡2或6（mod 8）。合并两组余数，n模1000只能为38、462、538、962；在316至999之间只有462、538、962，对应213444、289444、925444，共3个。若不接受“六位”这一手写范围，则把满足条件的平方根加上1000k仍保持末三位444，因此有无穷多个解。" }, "cross_reference_ids": [], "handwritten_additions": [ "题干上方红笔补写“六位”；印刷题干没有该字样。" ] }, { "id": "S13-Q012", "source_id": "S13", "source_filename": "26数论模块训练营第九讲解析(1).pdf", "lesson_or_test": "第9讲 三角形数与完全平方数", "source_question_label": "拓展1", "problem": "我们称能表示成1+2+3+⋯+k的形式的自然数为三角数。有一个四位数N，它既是三角数，又是完全平方数。则N=____。", "source_answer": "1225", "source_solution": "N=1+2+⋯+k=k(k+1)/2=a²；分k为偶数或奇数讨论，试探得到k=49，N=5²×7²=1225。", "topics": [ "三角形数", "完全平方数", "等差数列求和" ], "page_refs": { "problem": [ 5 ], "answer": [ 5 ], "solution": [ 5 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "1225。", "solution_status": "incomplete", "note": "检查32≤a≤99的全部四位平方，只有35²满足三角数判定；对应k=49。 来源对有限范围的“试探”未列完；本轮检验32²至99²全部候选，唯一为1225。", "verified_solution": "四位完全平方数写成a²，32≤a≤99。三角数条件等价于8a²+1为奇数平方（因为8T_k+1=(2k+1)²）。逐一检验这68个a值，只有a=35通过；于是N=35²=1225=1+2+…+49。" }, "cross_reference_ids": [] }, { "id": "S13-Q013", "source_id": "S13", "source_filename": "26数论模块训练营第九讲解析(1).pdf", "lesson_or_test": "第9讲 三角形数与完全平方数", "source_question_label": "拓展2", "problem": "有一个奇怪的四位数（首位不为0），它是完全平方数，它的数字和是完全平方数，用这个四位数除以它的数字和得到的结果还是完全平方数，并且它的因数个数还恰好等于它的数字和，那么也就是完全平方数。如果这个四位数的各位数字互不相同，那么这个四位数是多少？", "source_answer": "2601", "source_solution": "设四位数为n²。数字和≤30且为平方数，只考虑9、16、25；因数个数等于数字和且n²除以数字和仍为平方数。数字和25时因25整除且末两位需满足，源解析判不成立；数字和9时枚举3²×p²，候选1089、1521、2601、3249、4761、7569、8649，标记符合者为2601。", "topics": [ "完全平方数", "数字和", "因数个数", "质因数分解" ], "page_refs": { "problem": [ 5 ], "answer": [ 5 ], "solution": [ 5 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "2601。", "solution_status": "incomplete", "note": "独立穷举1000—9999验证唯一解为2601；其数字和9，商289=17²，正因数个数9。 来源候选筛选略去了若干排除理由；本轮补全因数个数奇偶与指数型分类。", "verified_solution": "数字和d至少6且因四位数各位不同最多30；d只能是9、16或25。完全平方数的正因数个数为奇数，所以d不可能是16。若d=25，N/25为平方数，故N末两位只能是00或25：00重复数字；25则前两位数字之和需为18，但两个不同数字的最大和为17，均不行。若d=9，因数个数为9且N为平方数，指数型只能为p^8或p²q²。N又含9；p^8分支只能是3^8=6561（数字和18，不符），另一分支为N=3²q²。四位范围内q只需查11、13、17、19、23、29、31；其中只有2601=3²×17²各位不同且数字和9。最后2601/9=289=17²，因数个数(2+1)(2+1)=9，故它满足全部条件。" }, "cross_reference_ids": [] }, { "id": "S13-Q014", "source_id": "S13", "source_filename": "26数论模块训练营第九讲解析(1).pdf", "lesson_or_test": "第9讲 三角形数与完全平方数", "source_question_label": "作业1", "problem": "有一堆粗细均匀的原木，堆成如图的形状。已知最上面一层有6根，共堆了25层。请问：这堆圆木共有多少根？", "source_answer": "450根", "source_solution": "第25层根数=6+25−1=30；总数=(6+30)×25÷2=450根。", "topics": [ "等差数列求和", "数列计数" ], "page_refs": { "problem": [ 7 ], "answer": [ 7 ], "solution": [ 7 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "450根。", "solution_status": "consistent", "note": "图示为三角形堆叠，底层30根；等差和为450。", "verified_solution": "共有25层。每下一层比上一层多1根，因此各层从上到下是6、7、…、30根。等差数列总和=(首项+末项)×项数÷2=(6+30)×25÷2=450。" }, "cross_reference_ids": [] }, { "id": "S13-Q015", "source_id": "S13", "source_filename": "26数论模块训练营第九讲解析(1).pdf", "lesson_or_test": "第9讲 三角形数与完全平方数", "source_question_label": "作业2", "problem": "在N=(x+10)+⋯⋯+(x+99)+(x+100)里，x是一个正整数，那么，能使N成为一个完全平方数的最小x值为多少？", "source_answer": "36", "source_solution": "N=(x+10+x+100)×91÷2=91(x+55)=7×13×(x+55)；(x+55)_min=7×13=91，所以x_min=91−55=36。", "topics": [ "等差数列求和", "完全平方数", "质因数分解" ], "page_refs": { "problem": [ 7 ], "answer": [ 7 ], "solution": [ 7 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "36。", "solution_status": "consistent", "note": "穷举正整数x并用平方判定，最小解x=36；更大的解不影响最小值。", "verified_solution": "从x+10加到x+100共91项，首末项和为2x+110，所以N=91(x+55)。因91=7×13没有平方因子，要使N成为平方数，x+55必须是91乘某个平方数，即x+55=91m²。m最小取1，得到x=36；代回N=91²=8281。" }, "cross_reference_ids": [] }, { "id": "S13-Q016", "source_id": "S13", "source_filename": "26数论模块训练营第九讲解析(1).pdf", "lesson_or_test": "第9讲 三角形数与完全平方数", "source_question_label": "作业3", "problem": "某一小组有四个工人，问起他们各自的年龄，其中一个人答道：‘我们四个人的年龄全都不一样，如果把我们的年龄相加，那一共是129岁；另外我们中有3个人的年龄是平方数，倒退15年，我们当中也有3个人的年龄是平方数。’他们四个人现在的年龄是____岁、____岁、____岁及____岁。", "source_answer": "16岁、24岁、25岁及64岁", "source_solution": "红笔解析：枚举平方数即可：16、24、25、64。", "topics": [ "完全平方数", "整数枚举", "年龄问题" ], "page_refs": { "problem": [ 7 ], "answer": [ 7 ], "solution": [ 7 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "16岁、24岁、25岁、64岁。", "solution_status": "incomplete", "note": "来源只写“枚举平方数”，未展示候选范围与唯一性；独立穷举年龄15—84内两类平方年龄，唯一四元组为16、24、25、64。", "verified_solution": "倒退15年后仍是实际年龄，故四人现在都至少15岁；总和129，所以每人不超过84岁。15—84之间，现在是平方数的年龄为16、25、36、49、64、81；倒退15年是平方数的年龄为15、16、19、24、31、40、51、64、79。检查这些候选中的四个不同年龄且总和129，唯一集合为{16,24,25,64}。其中现在为平方数的是16、25、64；15年前为平方数的是1、9、49，恰各有三人。" }, "cross_reference_ids": [] }, { "id": "S14-Q001", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "1", "problem": "三位数21□能被9整除，那么□内的数是____。", "source_answer": "6", "source_solution": "9的倍数特征是各数位上数字和是9的倍数。因此要保证2+1+□是9的倍数，故□内的数是6。", "topics": [ "整除性与倍数特征" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "6", "solution_status": "consistent", "note": "2+1+6=9。", "verified_solution": "9的倍数有个小窍门：各位数字相加后，也要是9的倍数。2+1=3，再加6刚好等于9，所以方框里填6。" }, "cross_reference_ids": [] }, { "id": "S14-Q002", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "2", "problem": "六位数95A14B是99的倍数，则两位数AB=____。", "source_answer": "62", "source_solution": "95A14B=950140+1000A+B=950140+990A+10A+B。由题意其为99的倍数，且10≤AB≤99，所以950150≤950140+10A+B≤950239；除以99得9597.47≤(950140+10A+B)/99≤9598.38，故商为9598，10A+B=62。", "topics": [ "整除性与倍数特征", "数字谜与数位分析" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "AB=62", "solution_status": "consistent", "note": "代入A=6、B=2得数为956142，且99×9658=956142，成立；来源推导中的中间表达式虽排版拥挤，结论正确。", "verified_solution": "把“AB”看成两位数10A+B。95A14B=950140+1000A+B=950140+990A+(10A+B)。因为990A一定是99的倍数，只要看950140+AB。AB在10到99之间，所以这个数在950150到950239之间。这里唯一的99倍数是950202=99×9598，于是AB=950202−950140=62。检查：A=6、B=2时，95A14B=956142=99×9658。" }, "cross_reference_ids": [] }, { "id": "S14-Q003", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "3", "problem": "四位数32□□能被23整除，则这个四位数最大是____，最小是____。", "source_answer": "", "source_solution": "试除法：设这个四位数是3299，则3299÷23=143……10，可得3299−10=3289，3289是23的倍数，且是能被23整除的最大的四位数。设这个四位数是3200，则3200÷23=139……3，可得3200+(23−3)=3220，3220是23的倍数，且是能被23整除的最小的四位数。", "topics": [ "整除性与倍数特征", "范围与最值" ], "page_refs": { "problem": [ 1 ], "answer": [], "solution": [ 1 ] }, "ambiguities": [ "第1页第3题未见独立的【答案】标记，答案由解析明确给出。" ], "duplicate_notes": "", "review": { "answer_status": "verified_from_solution", "verified_answer": "最大3289；最小3220。", "solution_status": "consistent", "note": "来源未单列答案栏；解析给出的最大值3289=23×143、最小值3220=23×140，均在3200～3299范围。", "verified_solution": "四位数32□□在3200到3299之间。最大值：3299÷23余10，所以从3299减去10，得到3289=23×143。最小值：3200÷23余3，再加上23−3=20，得到3220=23×140。" }, "cross_reference_ids": [ "XR-01" ] }, { "id": "S14-Q004", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "4", "problem": "如果一天所在的月份是质数，日期是质数，月份和日期的和也是质数，那么我们称这样的日子为“质数日”。一年最多有____个“质数日”。", "source_answer": "8", "source_solution": "两个质数的和是质数，其中必有一个是2，另一个是奇质数，可以是3、5、11、17、29。因此最多有8个质数日。", "topics": [ "质数与合数", "分类枚举" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "8个（在闰年达到上限）", "solution_status": "incomplete", "note": "闰年2月可取质数日3、5、11、17、29共5个；3、5、11月各可取日期2，共3个，合计8。平年缺少2月29日，最多7个；原卷解析未明确说明闰年边界，已补独立说明。", "verified_solution": "如果两个质数相加还是质数，其中一个必须是2；否则两个奇质数相加是大于2的偶数，不会是质数。闰年2月的质数日期3、5、11、17、29都可以，日期加2后仍是质数，共5天。若日期是2，月份可选3、5、11（因为7+2=9不是质数），共3天。最多5+3=8天；平年没有2月29日，最多7天。" }, "cross_reference_ids": [] }, { "id": "S14-Q005", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "5", "problem": "N为一个非0整数，使180x=N²成立的最小自然数，则x=____。", "source_answer": "5", "source_solution": "180=2×2×3×3×5，x最小为5。", "topics": [ "平方数与完全平方数", "质因数分解" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "5", "solution_status": "incomplete", "note": "180=2²×3²×5；为使乘积成为平方数，因数5需再补一个5，因此最小x=5且180×5=30²。来源只给质因数分解和答案，已补足最小性理由。", "verified_solution": "180=2²×3²×5。要让180×x成为平方数，每种质因数都要成对出现；2和3已经成对，5还单独剩一个，所以x最少要补一个5。取x=5时，180×5=900=30²，因此最小x=5。" }, "cross_reference_ids": [] }, { "id": "S14-Q006", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "6", "problem": "已知A、B两数的最小公倍数是360，最大公因数是6，若A=30，则B=____。", "source_answer": "72", "source_solution": "根据最小公倍数×最大公因数=A×B，知道B=360×6÷30=72。", "topics": [ "最大公因数与最小公倍数" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "72", "solution_status": "consistent", "note": "30×72=2160=360×6。", "verified_solution": "两数相乘=最大公因数×最小公倍数，所以30×B=6×360，B=6×360÷30=72。检查：gcd(30,72)=6，lcm(30,72)=360。" }, "cross_reference_ids": [] }, { "id": "S14-Q007", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "7", "problem": "已知一个苹果重4/15千克，一个梨重5/24千克，且苹果和梨的总重量相同，最少有____个苹果。（水果为整数个）", "source_answer": "25", "source_solution": "本题实质上是求4/15和5/24的最小公倍数，[4/15，5/24]=[4,5]/[15,24]=20/3，因此苹果个数是20/3÷4/15=25个。", "topics": [ "分数与最小公倍数", "比例与应用题" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "25个苹果", "solution_status": "consistent", "note": "最少苹果数为25；对应梨数为32（此前复核备注误写16，已更正）。两种水果的总重量均为20/3千克。", "verified_solution": "设苹果有a个、梨有b个。重量相同，所以4a/15=5b/24。两边乘120，得到32a=25b。25和32没有共同因数，因此最小的正整数解是a=25、b=32。检查：25个苹果重4×25/15=20/3千克，32个梨也重5×32/24=20/3千克。题目只问苹果数，答案是25个。" }, "cross_reference_ids": [] }, { "id": "S14-Q008", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "8", "problem": "180一共有____个因数；这些因数的和是____。", "source_answer": "18，546", "source_solution": "因为180=2²×3²×5，180的因数个数为(2+1)×(2+1)×(1+1)=18（个），180的因数的和为(1+2¹+2²)×(1+3¹+3²)×(1+5¹)=546。", "topics": [ "质因数分解", "因数个数与因数和" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "18个因数；因数和546", "solution_status": "consistent", "note": "因数个数公式和约数和公式计算均正确。", "verified_solution": "180=2²×3²×5。一个因数可从2、3、5的指数中分别选择，所以因数个数=(2+1)(2+1)(1+1)=18。所有因数的和=(1+2+4)(1+3+9)(1+5)=7×13×6=546。" }, "cross_reference_ids": [] }, { "id": "S14-Q009", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "9", "problem": "987654除以5、7、11，余数依次为____、____、____。", "source_answer": "4，3，8", "source_solution": "利用余数特征很快可得出分别为4、3、8。", "topics": [ "同余与余数", "整除性与倍数特征" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "除以5余4；除以7余3；除以11余8", "solution_status": "incomplete", "note": "逐一计算余数为4、3、8。", "verified_solution": "987654=5×197530+4，所以除以5余4；987654=7×141093+3，所以除以7余3；987654=11×89786+8，所以除以11余8。" }, "cross_reference_ids": [] }, { "id": "S14-Q010", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "10", "problem": "解同余方程：6x≡3x+7(mod11)，x≡____(mod11)。", "source_answer": "6", "source_solution": "6x−3x≡7(mod11)，3x≡7(mod11)，3x≡7+11≡18(mod11)，x≡6(mod11)。", "topics": [ "同余方程", "模运算" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "x≡6 (mod 11)", "solution_status": "incomplete", "note": "3×6=18≡7 (mod 11)。", "verified_solution": "6x≡3x+7 (mod 11)，两边减去3x，得到3x≡7 (mod 11)。因为3×4=12≡1 (mod 11)，所以3的逆元是4；两边乘4，x≡28≡6 (mod 11)。" }, "cross_reference_ids": [] }, { "id": "S14-Q011", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "11", "problem": "2020²⁰²¹+2021²⁰²⁰除以7的余数是____。", "source_answer": "4", "source_solution": "", "topics": [ "同余与余数", "幂的周期性" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "余数为 4。", "solution_status": "not_provided", "note": "独立复算：2020≡4、2021≡5 (mod 7)；4^2021≡2、5^2020≡2，和为4。来源答案4正确，原资料未附解析。此前核验把两个底数的余数各少算1，已撤销误报。", "verified_solution": "2020≡4、2021≡5 (mod 7)。4³≡1，因此 4^2021≡4²≡2；5³≡−1，因此 5^2020=(5³)^673·5≡(−1)^673·5≡−5≡2。两项相加余数为 4。原资料未附解析。" }, "cross_reference_ids": [] }, { "id": "S14-Q012", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "12", "problem": "442，297，210分别除以某个大于1的自然数，能得到相同的余数，则该自然数是____。", "source_answer": "29", "source_solution": "442−297=145，442−210=232，297−210=87；145=5×29，232=8×29，87=3×29，则这个余数为29。", "topics": [ "同余与余数", "最大公因数" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "除数为29；共同余数为7", "solution_status": "error", "note": "d=29是满足条件的除数；三数除以29的共同余数是7。原解析最后一句把除数29误称为余数，已标记来源解法有误并保留原文。", "verified_solution": "若442、297、210除以d后余数相同，那么它们两两相减都能被d整除。差分别是145、232、87，它们的最大公因数是29。因为d>1，只能取d=29。检查：442=29×15+7，297=29×10+7，210=29×7+7；因此除数是29，共同余数是7。原解析最后把“除数29”误写成“余数29”，原文保留在source_solution中。" }, "cross_reference_ids": [] }, { "id": "S14-Q013", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "13", "problem": "六一儿童节，幼儿园买来一些气球（不超过100个），如果平均分给10位小朋友，最后一位小朋友就会少1个；如果平均分给12位小朋友，最后一位小朋友还会少1个；如果平均分给14位小朋友，最后一位小朋友就会多3个。幼儿园共买来____个气球。", "source_answer": "59", "source_solution": "100以内14的倍数有14、28、42、56、70、84、98；分别加3得17、31、45、59、73、87、101，满足分给12个小朋友最后一位少1个的有12×5−1=59，满足分给10个小朋友最后一位少1个的有6×10−1=59，则幼儿园共买来59个气球。", "topics": [ "同余与余数", "不定方程与应用题" ], "page_refs": { "problem": [ 1 ], "answer": [ 1 ], "solution": [ 1 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "59", "solution_status": "error", "note": "原解析把101列入“不超过100”的候选，又漏掉候选3；完整列出N≤100且N≡3 (mod14)的数后，仍只有59同时满足另外两个余数条件。答案59正确，来源中间枚举有误。", "verified_solution": "把气球数记作N。按14人分配时“多3个”，所以N除以14余3。又N≤100，因此候选是3、17、31、45、59、73、87。按12人分配“少1个”，表示N除以12余11；候选中只有59满足。再检查59除以10余9，正好符合10人分配时少1个。因此唯一答案是59。" }, "cross_reference_ids": [] }, { "id": "S14-Q014", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "14", "problem": "有A、B两个整数，A的各位数字之和为36，B的各位数字之和为25，且两数相加时进位三次，那么A+B的各位数字之和是____。", "source_answer": "", "source_solution": "进三次位数字和少3×9=27，则36+25−27=34。", "topics": [ "数字和与数位分析", "加法进位" ], "page_refs": { "problem": [ 1 ], "answer": [], "solution": [ 2 ] }, "ambiguities": [ "第1页题面后未见独立答案；解析位于第2页页首。" ], "duplicate_notes": "", "review": { "answer_status": "verified_from_solution", "verified_answer": "34。", "solution_status": "consistent", "note": "来源未单列答案栏；三次进位使数位和总计减少3×9，36+25−27=34。", "verified_solution": "每发生一次进位，个位数字少10、左边一位多1，数字和合起来少9。进位三次，数字和共少3×9=27。A+B的数字和是36+25−27=34。" }, "cross_reference_ids": [] }, { "id": "S14-Q015", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "15", "problem": "一个数的平方数有5个因数，那么这个数的立方有____个因数。", "source_answer": "7", "source_solution": "其形式必为a⁴，其中a为质数，它的立方为a⁶，有7个因数。", "topics": [ "因数个数与因数和", "质因数分解", "幂的运算" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "7", "solution_status": "consistent", "note": "平方的因数个数为5意味着平方为p⁴；原数为p²，立方为p⁶，因数数为7。", "verified_solution": "一个正整数平方的因数个数是5。因数个数为5时，平方只能是某个质数p的四次方p⁴，所以原数是p²。它的立方是p⁶，因数个数为6+1=7。" }, "cross_reference_ids": [] }, { "id": "S14-Q016", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "16", "problem": "一个四位数abcd和它的反序数dcba都是65的倍数，这个四位数abcd是____。", "source_answer": "5005", "source_solution": "abcd≡dcba≡abcd+dcba≡abcd−dcba≡0(mod65)。由65=13×5，得5|(a+d)，13|(b+c)，5|(a−d)（令a>d）。由5|(a+d)可得a+d=10，由5|(a−d)可得a−d=0或5，只有a+d=10、a−d=0有整数解，故a=d=5。又13|(b+c)，可得b+c=13或0。若b+c=13，则abcd=5005+90b+130≡25b≡0(mod65)，b=0、c=13不符合；若b+c=0，则b=c=0，abcd=5005≡0(mod65)，符合，故abcd=5005。", "topics": [ "整除性与倍数特征", "同余与余数", "数字谜与数位分析" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [ "解析中‘令a>d’省略了a=d等边界讨论，但最终答案在四位反序数约束下成立。" ], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "5005", "solution_status": "error", "note": "原解析称“令a>d”后又推到a=d=5，前提与结论矛盾；独立穷尽证明确认答案唯一为5005。", "verified_solution": "设四位数为abcd，反序数dcba也是四位数，所以a、d都不是0。两数都是65的倍数，特别都能被5整除；abcd的末位d和dcba的末位a因此只能是5（不能是0），故a=d=5。于是abcd=5005+100b+10c，反序数=5005+100c+10b。5005已能被65整除；余下条件等价于13整除10b+c，并且13整除10c+b。相减得9(b−c)能被13整除；因9与13没有公因数，b−c能被13整除。b、c都是0到9的数字，所以只能b=c。代回后13整除11b；11与13没有公因数，因此b=0，c=0。唯一数是5005；检查5005=65×77，反序仍是5005。原解析称“令a>d”后又取a=d，前提互相矛盾，独立证明已替代该推导。" }, "cross_reference_ids": [] }, { "id": "S14-Q017", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "17", "problem": "求出所有正整数n，使得25+n能整除25×n。", "source_answer": "见解析", "source_solution": "由题意25n/(25+n)=m（m为自然数），于是1/25+1/n=1/m，即1/25=1/m−1/n。原页再列出1/25=5/(25×4)−1/(25×4)=1/20−1/100，以及1/25=25/(25×24)−1/(25×24)=1/24−1/600，给出n=100、600。", "topics": [ "整除性与倍数特征", "分式与不定方程" ], "page_refs": { "problem": [ 2 ], "answer": [ 2 ], "solution": [ 2 ] }, "ambiguities": [], "duplicate_notes": "", "review": { "answer_status": "verified", "verified_answer": "n=100、600。", "solution_status": "incomplete", "note": "用D=n+25可化为D|625且D>25，故D只可能125或625，得到全部n=100、600；来源只列两个分数分解，未说明穷尽性。", "verified_solution": "设D=n+25。因为n是正整数，所以D>25。若D能整除25n，那么它也能整除25(n+25)−25n=625。625=5⁴，且大于25的正因数只有125和625。因此D=125或625，n=D−25，得到n=100或600。代回原式，两者均满足；除此以外没有其他因数可选。" }, "cross_reference_ids": [] }, { "id": "S14-Q018", "source_id": "S14", "source_filename": "25数论训练营综合测试解析版(1).pdf", "lesson_or_test": "数论训练营 综合测试", "source_question_label": "18", "problem": "已知等式1/15=1/a+1/b，其中a、b是非零自然数，求a+b的所有可能取值。", "source_answer": "", "source_solution": "易知1/15=(m+n)/(15(m+n))=m/[15(m+n)]+n/[15(m+n)]。令(m,n)为互质的一对数，现在要让分子为1，只需m、n是15的一对互质的因数即可。当(m,n)=(1,1)时，1/15=1/30+1/30，此时a+b=60；当(m,n)=(1,3)时，1/15=1/60+1/20，此时a+b=80；当(m,n)=(1,5)时，1/15=1/90+1/18，此时a+b=108；当(m,n)=(1,15)时，1/15=1/240+1/16，此时a+b=256；当(m,n)=(3,5)时，1/15=1/40+1/24，此时a+b=64。", "topics": [ "分式与不定方程", "互质与因数分解", "埃及分数" ], "page_refs": { "problem": [ 2 ], "answer": [], "solution": [ 2 ] }, "ambiguities": [ "题面写‘非零自然数’，按解析实际按正整数处理；若自然数包含0，因分母不能为0仍不影响。第2页未见独立【答案】标记。" ], "duplicate_notes": "", "review": { "answer_status": "verified_from_solution", "verified_answer": "a+b∈{60,64,80,108,256}。", "solution_status": "incomplete", "note": "完整分类为(a−15)(b−15)=225的无序因数对；来源列出了五个和，但未充分说明因数对已经穷尽。", "verified_solution": "由1/15=1/a+1/b，两边乘15ab，得ab=15a+15b，也就是(a−15)(b−15)=225。右边是正数，所以两个括号同号。a、b为正整数，若两括号都为负，它们的绝对值最多14，乘积最多196，不可能等于225；因此两括号都为正。225的无序正因数对是(1,225)、(3,75)、(5,45)、(9,25)、(15,15)。分别加回15，得到(a,b)=(16,240)、(18,90)、(20,60)、(24,40)、(30,30)，所以a+b的所有可能值为256、108、80、64、60，即{60,64,80,108,256}。" }, "cross_reference_ids": [ "XR-05" ] } ], "indexes": { "lesson_test": [ { "source_id": "S01", "lesson_or_test": "数论方法训练营 第1讲 整除符号与性质", "question_ids": [ "S01-Q001", "S01-Q002", "S01-Q003", "S01-Q004", "S01-Q005", "S01-Q006", "S01-Q007", "S01-Q008", "S01-Q009", "S01-Q010", "S01-Q011", "S01-Q012", "S01-Q013", "S01-Q014", "S01-Q015", "S01-Q016" ] }, { "source_id": "S02", "lesson_or_test": "数论方法训练营第2讲：质合与构造论证", "question_ids": [ "S02-Q001", "S02-Q002", "S02-Q003", "S02-Q004", "S02-Q005", "S02-Q006", "S02-Q007", "S02-Q008", "S02-Q009", "S02-Q010", "S02-Q011", "S02-Q012", "S02-Q013", "S02-Q014", "S02-Q015", "S02-Q016", "S02-Q017", "S02-Q018", "S02-Q019" ] }, { "source_id": "S03", "lesson_or_test": "数论方法训练营第3讲 唯一分解定理", "question_ids": [ "S03-Q001", "S03-Q002", "S03-Q003", "S03-Q004", "S03-Q005", "S03-Q006", "S03-Q007", "S03-Q008", "S03-Q009", "S03-Q010", "S03-Q011", "S03-Q012", "S03-Q013", "S03-Q014", "S03-Q015", "S03-Q016", "S03-Q017" ] }, { "source_id": "S04", "lesson_or_test": "数论短期班 小测1", "question_ids": [ "S04-Q001", "S04-Q002", "S04-Q003", "S04-Q004" ] }, { "source_id": "S05", "lesson_or_test": "数论模块班小测2", "question_ids": [ "S05-Q001", "S05-Q002", "S05-Q003", "S05-Q004", "S05-Q005", "S05-Q006", "S05-Q007" ] }, { "source_id": "S06", "lesson_or_test": "第4讲 分解与分数拆分（例题精讲、练习、拓展提升、课后巩固）", "question_ids": [ "S06-Q001", "S06-Q002", "S06-Q003", "S06-Q004", "S06-Q005", "S06-Q006", "S06-Q007", 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"question_ids": [ "S12-Q020" ] }, "分式与不定方程": { "question_count": 2, "question_ids": [ "S14-Q017", "S14-Q018" ] }, "分数与最小公倍数": { "question_count": 1, "question_ids": [ "S14-Q007" ] }, "分数拆分（埃及分数）": { "question_count": 1, "question_ids": [ "S11-Q001" ] }, "分数方程": { "question_count": 3, "question_ids": [ "S06-Q009", "S06-Q010", "S06-Q015" ] }, "分数的最大公因数": { "question_count": 1, "question_ids": [ "S08-Q004" ] }, "分数的最大公因数与最小公倍数": { "question_count": 3, "question_ids": [ "S03-Q012", "S03-Q013", "S03-Q014" ] }, "分数的最小公倍数": { "question_count": 1, "question_ids": [ "S08-Q004" ] }, "分数约分": { "question_count": 1, "question_ids": [ "S03-Q012" ] }, "分类枚举": { "question_count": 2, "question_ids": [ "S10-Q017", "S14-Q004" ] }, "分类讨论": { "question_count": 1, "question_ids": [ "S03-Q011" ] }, "分解模数": { "question_count": 1, "question_ids": [ "S09-Q012" ] }, "加法进位": { "question_count": 1, "question_ids": [ "S14-Q014" ] }, "单位分数方程": { "question_count": 1, "question_ids": [ "S06-Q011" ] }, "参数余数与代数恒等式": { 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"question_ids": [ "S10-Q014", "S11-Q005", "S14-Q010" ] }, "同余方程与周期计数": { "question_count": 1, "question_ids": [ "S07-Q008" ] }, "同余方程组": { "question_count": 7, "question_ids": [ "S09-Q005", "S09-Q006", "S09-Q007", "S09-Q008", "S09-Q009", "S09-Q014", "S09-Q015" ] }, "同余法求正整数解": { "question_count": 2, "question_ids": [ "S09-Q003", "S09-Q004" ] }, "同余的加减乘运算": { "question_count": 2, "question_ids": [ "S07-Q001", "S07-Q003" ] }, "周期性": { "question_count": 1, "question_ids": [ "S07-Q010" ] }, "周期性与等比数列同余": { "question_count": 1, "question_ids": [ "S07-Q004" ] }, "周期性与计数": { "question_count": 1, "question_ids": [ "S01-Q013" ] }, "唯一分解定理": { "question_count": 1, "question_ids": [ "S03-Q001" ] }, "四位数排序": { "question_count": 1, "question_ids": [ "S12-Q018" ] }, "因式分解": { "question_count": 7, "question_ids": [ "S01-Q013", "S06-Q002", "S06-Q003", "S06-Q005", "S06-Q007", "S08-Q005", "S13-Q010" ] }, "因式分解与分类枚举": { "question_count": 1, "question_ids": [ "S06-Q011" ] }, "因式分解与因数个数": { 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"question_ids": [ "S12-Q017" ] }, "完全平方数": { "question_count": 14, "question_ids": [ "S06-Q016", "S13-Q001", "S13-Q002", "S13-Q003", "S13-Q004", "S13-Q005", "S13-Q007", "S13-Q008", "S13-Q009", "S13-Q011", "S13-Q012", "S13-Q013", "S13-Q015", "S13-Q016" ] }, "完全立方数": { "question_count": 2, "question_ids": [ "S13-Q007", "S13-Q008" ] }, "带余除法": { "question_count": 2, "question_ids": [ "S04-Q003", "S08-Q002" ] }, "带分数化假分数": { "question_count": 1, "question_ids": [ "S03-Q014" ] }, "带参数余数问题": { "question_count": 1, "question_ids": [ "S07-Q006" ] }, "带参数整除": { "question_count": 4, "question_ids": [ "S01-Q009", "S01-Q011", "S01-Q012", "S01-Q016" ] }, "幂的分解与奇偶性": { "question_count": 1, "question_ids": [ "S06-Q016" ] }, "幂的同余": { "question_count": 4, "question_ids": [ "S09-Q010", "S09-Q011", "S09-Q012", "S11-Q004" ] }, "幂的周期性": { "question_count": 5, "question_ids": [ "S07-Q008", "S07-Q009", "S07-Q011", "S07-Q014", "S14-Q011" ] }, "幂的整数结构": { "question_count": 1, "question_ids": [ "S06-Q003" ] }, 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"S14-Q018" ] }, { "family": "分数、分式与埃及分数", "topic_count": 7, "topics": [ "分数拆分（埃及分数）", "分数方程", "分数约分", "单位分数方程", "埃及分数", "埃尔德什-施特劳斯型分数方程", "带分数化假分数" ], "question_count": 9, "question_ids": [ "S03-Q012", "S03-Q014", "S06-Q008", "S06-Q009", "S06-Q010", "S06-Q015", "S06-Q011", "S11-Q001", "S14-Q018" ] }, { "family": "数位、位值与进制", "topic_count": 37, "topics": [ "不进位数字构造", "九进制", "二进制", "五进制", "位值制与数的拼接", "位值原理", "八进制", "六进制转换", "反序数", "多进制表示", "多进制转换", "字母代数与进制", "幂的末位数字", "数字倒置与最值", "数字和", "数字和与倍乘进位", "数字和与减法借位", "数字和与加法进位", "数字和与数位分析", "数字和与整式加法进位", "数字和与模9", "数字和判定法", "数字平方和", "数字排列", "数字排列与三位数和", "数字方程", "数字根", "数字谜", "数字谜与数位分析", "数字选择与计数", "数码和", "数码和与进位关系", "模5与末位数字", "模7与末位数字", "进制方程", "进制运算", "重复数" ], "question_count": 55, "question_ids": [ "S01-Q001", "S01-Q002", "S01-Q003", "S01-Q005", "S01-Q006", "S01-Q007", "S01-Q008", "S13-Q013", "S02-Q009", "S02-Q010", "S02-Q016", "S02-Q011", "S02-Q019", "S12-Q018", "S02-Q018", "S05-Q001", "S07-Q003", "S07-Q004", "S07-Q009", "S07-Q010", 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"奇偶性", "平均数", "指数分拆", "指数变化", "指数方程", "排除法", "整数分解", "整数因式分解", "整数方程", "整数枚举", "星期与模7", "最大值与最小值", "最大值问题", "最大长度", "最小值问题", "最小正整数", "末三位问题", "枚举与排除", "模逆元", "正整数通解与最小值", "比例", "竖式加法进位", "范围内整数解", "试除法", "辗转相除法", "连续偶数", "连续奇数", "连续自然数" ], "question_count": 56, "question_ids": [ "S01-Q005", "S01-Q011", "S01-Q014", "S01-Q008", "S08-Q003", "S09-Q009", "S01-Q013", "S06-Q002", "S06-Q003", "S06-Q005", "S06-Q007", "S08-Q005", "S13-Q010", "S01-Q015", "S02-Q004", "S02-Q008", "S02-Q013", "S02-Q015", "S02-Q017", "S02-Q018", "S03-Q002", "S03-Q003", "S03-Q004", "S06-Q013", "S03-Q007", "S05-Q003", "S05-Q004", "S03-Q008", "S03-Q009", "S03-Q017", "S03-Q011", "S03-Q015", "S12-Q022", "S03-Q016", "S05-Q006", "S06-Q011", "S07-Q011", "S09-Q002", "S09-Q013", "S09-Q012", "S09-Q014", "S09-Q015", "S10-Q001", "S10-Q002", "S10-Q015", "S10-Q003", "S10-Q004", "S10-Q007", "S10-Q017", "S14-Q004", "S11-Q005", "S12-Q006", "S12-Q018", "S13-Q008", "S13-Q016", "S14-Q014" ] } ] }, "cross_references": [ { "id": "XR-01", "relationship": "exact_prompt_duplicate", "question_ids": [ "S01-Q015", "S14-Q003" ], "note": "“四位数32□□能被23整除，求最大值和最小值”题面相同；分别出现在第1讲作业与综合测试，保留两条来源记录。" }, { "id": "XR-02", "relationship": "exact_prompt_duplicate", "question_ids": [ "S02-Q010", "S02-Q016" ], "note": "第2讲练习4与拓展2题面和答案相同，保留两个原始题号并标为重复。" }, { "id": "XR-03", "relationship": "repeated_across_pages", "question_ids": [ "S06-Q009" ], "note": "第4讲练习3在PDF第2、3页重复出现，抽取为一条记录，页码映射保留重复页。" }, { "id": "XR-04", "relationship": "shared_identity_variant", "question_ids": [ "S13-Q003", "S13-Q004" ], "note": "第9讲例题2与练习2复用1234567654321×49=7777777²这一恒等式，问法不同，属于变式/知识重叠，不按重复题合并。" }, { "id": "XR-05", "relationship": "related_variant", "question_ids": [ "S06-Q008", "S14-Q018" ], "note": "第4讲例题3与综合测试第18题都围绕1/15=1/a+1/b；一个求a+b最大值，一个求所有可能值，属于相关变式。" }, { "id": "XR-06", "relationship": "same_method_different_data", "question_ids": [ "S07-Q005", "S07-Q006", "S07-Q012" ], "note": "第5讲例题3、练习3、作业1均为相同余数/差值模型，但参数不同，不是重复题。" }, { "id": "XR-07", "relationship": "same_method_different_data", "question_ids": [ "S10-Q001", "S10-Q002", "S10-Q015" ], "note": "第7讲例题1、练习1、作业1均属互不相同面值的不可表示数/线性组合模型，保留为不同题。" } ] }